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      <title>Ch.5 - Energy by Maissa Jenblat</title>
      <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx</link>
      <description>Share your notes for every page </description>
      <language>en-us</language>
      <pubDate>2017-10-11 04:46:29 UTC</pubDate>
      <lastBuildDate>2025-11-26 05:50:09 UTC</lastBuildDate>
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      <item>
         <title>Hiba EL Chafie</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091897</link>
         <description><![CDATA[<div>Page 1 observation:<br>-The applied force and the spring constant  forces are equal but in opposite directions.<br>-knowing that the units for k(constant of elasticity) in N/m.<br>- when we increased the k the less stretchy/more stiff the spring became.<br>-The relationship between the between the applied force and the displacement vector for k were linear and what determines the elasticity is the k. <br>Page 2 Observation: <br>-  Fs=-Kx <br>- the negative k indicates that the spring force is in the opposite direction of the displacement. the restoring force and the x are directly proportional but in opposite directions and thats why to find k its negative. <br><br>Page 3 Observation: <br>- To find k w use the equation k=Fs/-x and with the calculations it showed that the k is 9.8N/m.<br>- The more the mass the more elongation and higher spring force. <br>- </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:53:20 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091897</guid>
      </item>
      <item>
         <title>Hanin H</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091927</link>
         <description><![CDATA[<div><strong>Page 1: <br></strong>Applied force and spring (restoring) force are in opposite directions in equal magnitudes. The spring constant, or constant of elasticity, abbreviated k, is a measure of force over displacement, more commonly Newtons/meter. The higher the value of k, the stiffer the spring, vice versa. The relationship between applied force and displacement is linear because they increase and decrease proportionally to one another. <br><br><strong>Page 2: <br></strong>Physicist Robert Hooke discovered that most solids, under heavy loads, behave elastically. Even the ones that do not have elastic properties end up showing elasticity. He derived the formula<br><br><em>Fs = -kx<br></em><br><em>Fs</em> stands for spring force (N)<br><em>k</em> stands for the constant of elasticity (N/m)<br><em>x</em> stands for the elongation or deformation; the difference in length of spring<br><br>This equation is in the form y = mx +b where k is the slope and b is 0. The spring force is dependent on the k.The negative sign is there to qualitatively show that the direction of the force and displacement are in opposite directions.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:53:38 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091927</guid>
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      <item>
         <title></title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091939</link>
         <description><![CDATA[<div>Page 1: <br>For applied force </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:53:43 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091939</guid>
      </item>
      <item>
         <title>Zarish Zafar</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091941</link>
         <description><![CDATA[<div>Page 1: The size of the applied force and the restoring force are equal; however, their directions are opposite. Also, the unit of k, the spring constant, is N/m because the spring constant depends on the applied force and the displacement of the spring. Moreover, a stiffer/less stretchy  spring has a higher value of k as more energy is needed to stretch or compress the spring. The applied force and the displacement of the spring have a linear relationship because when the force increases so does the displacement of the spring. In this case, the k is kept constant and does not change. <br><br>Page 2: Robert Hooke said that most solid beave (at times) with elastic properties, even very inelastic materials behave elastically under large loads. Thus, he proposed Hooke's law. Hooke's law is Fs= -kx, Fs is the restoring force and it is in units of Newton. k is the constant of elasticity and it describes the elasticity of a spring. x is the elongation or deformation of the spring, in other words the displacement of the spring. The negative sign means that the spring force is in the opposite direction of the displacement of the spring. <br><br>Page 3: </div>]]></description>
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         <pubDate>2020-02-04 05:53:45 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091941</guid>
      </item>
      <item>
         <title>Abbas Al-Howaidi</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091950</link>
         <description><![CDATA[<div>Key terms<br>elongation x deformation<br>spring force<br>restoring force<br>elasticity <br><br>Pg-1<br>      In the simulation I was able to identify that the spring force and the applied force are going in opposite direction but are equal in magnitude. The spring constant is k which can be measures in N/m, this k identifies the elasticity of this state, the more of k we have the stiffer the spring becomes.You can use k to calculate the spring force F=-kx, the negative sign is the spring force in the opposite direction of displacement.<br>Pg-2<br>     The relationship of the applied force and the displacement is linear(the applied force increases as the displacement also increases) when k is held constant. <br>     </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:53:49 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091950</guid>
      </item>
      <item>
         <title>Ibrahim Majid</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091973</link>
         <description><![CDATA[<div>observation  of page 1 : <br> we observed that force  from the spring  and k value , effect the direction and size of applied force. further observation of playing around with the blue and red slider which controlled the applied force and k value, we found out that when  the applied force is kept constant and k is increased the spring becomes less elastic more stiff. we also found that the relation ship between the applied force and displacement is in a linear relationship.<br><br>page 2 :  in page 2 we observe that because even inelastic materials can behave elastically under a heavy load,  we need a equation to show that this is true. this is defined by Fs=-kx in which Fs is the spring force measured in newtons, the k is the spring constant, measured in N/M, and x is deformation of spring.  when 2 springs are working consecutively they double the deformation given the k and Fs, and when parallel to each other they remain he same.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:07 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091973</guid>
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      <item>
         <title></title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091976</link>
         <description><![CDATA[
Hiba EL Cga
Your avatar
Add comment
]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:09 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091976</guid>
      </item>
      <item>
         <title></title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091979</link>
         <description><![CDATA[
Hiba EL C
Your avatar
Add comment
]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:10 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091979</guid>
      </item>
      <item>
         <title>Hooke&#39;s Law - Abdel Raheem Naji</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091982</link>
         <description><![CDATA[<div>pg1) The force of a spring is the force exerted by a stretched or compressed spring upon another object it is attached to. The magnitude of the spring force is equal to that of the applied force, but are opposes it in direction. The unit for the spring constant, abbreviated as k, is N/m as it depends on the force applied over the distance the spring is stretched. As k increases, the spring becomes stiffer/less stretchy. This is because the force needed to stretch or compress the spring a specific distance increases. After some experimentation, one can come to the conclusion that the relationship between applied force and displacement is linear, not quadratic. They are directly proportional as long as k (spring constant) remains the same.<br>pg2) Fs is the spring force, or restoring force, and is measured in Newtons. The equation to find the spring force is Fs=-kx. However, the (-) negative sign is only conceptual and is disregarded when problem solving. The negative sign is used only to show that the force of spring is in the opposite direction of the displacement</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:12 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091982</guid>
      </item>
      <item>
         <title>Zina Alrahhal </title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091991</link>
         <description><![CDATA[<div>-Force needed to stretch the spring is directly proportional to the displacement (Fp= K.x) until equilibrium<br>-Magnitude to the restoring force is proportional to how much you stretch or compress the string from its natural length. <br>Spring constant (k): how much force needed to stretch a spring by one meter (N/m) (Fs= -k.x) <br>The higher value of spring constant, the more stiffness it has. <br>The negative sign in the equation is only used to indicate that the direction of the restoring force is opposite to the direction of the force applied. <br>Compressing the spring causes it to push out, and stretching it causes the spring to pull in,  as the spring will be tempted to reach equilibrium. <br>when the spring reaches equilibrium, x=0. <br>The graph of spring force/ elongation (k/x) will be linear because as the spring force increases, the displacement increases. Increase in mass results in an increase of force, since F=mg.<br><br></div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:16 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440091991</guid>
      </item>
      <item>
         <title>Amna Shah</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092009</link>
         <description><![CDATA[<div>Page 1: The direction of both the forces, applied and spring, are in equally opposite directions but they have the same size. The spring constant is also known as the constant of elasticity which is always measured in n/m. It measures the stiffness of the spring. When the value of k is higher, the force is stronger. Thus, the stiffer and less stretchy it is. <br>There is a unique relationship between the applied force and the displacement for a constant k, which makes it linear. Since the values decrease or increase by the same rate each time. When the force is increased, the displacement also increases.<br><br>Page 2: Hooke's law is Fs=-kx, the negative sign indicates that the displacement is in the opposite direction of the spring force, it predicts the direction of the force.The spring force is a restoring force that returns the spring to its original position.  The x is used to measure the elongation of the the spring. The spring force and the constant of elasticity are proportional, but the elasticity constant (k) determines the force. </div>]]></description>
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         <pubDate>2020-02-04 05:54:26 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092009</guid>
      </item>
      <item>
         <title>MOAAZ</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092065</link>
         <description><![CDATA[<div>Page:1<br>The applied force and the spring force are in opposite direction and they have a same size.<br>The units of the constant of elasticity is N/m(K).<br>as the value of K increases the spring becomes stiffer and less stretchy.<br>As the applied force increases, the displacement increases if k is held constant. This results in a linear relationship.<br><br>Page2: <br>Hooke discovered a relationship between k(spring constant), Fs(the spring force), and X(the elongation or deformation of the spring).  Fs=-kx, the negative is a conceptual sign that means that the spring force opposes the displacement of the spring;therefore, the slope is negative.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:54:55 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092065</guid>
      </item>
      <item>
         <title>Ahmed Mohamed</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092077</link>
         <description><![CDATA[<div>page 1: <br>The size of applied force and the spring force are at equal length. The applied force and the spring force are opposite in direction . Constant elasticity is k and having a higher k value makes the spring stiffer. The relationship of the applied force and the displacement vector when having a constant k, they are linear to each other.<br><br>Page 2:<br>Hooke established a law to show the correlation between Spring force Fs, the constant elasticity k, and the elongation or deformation x with Fs=-kx. The negative sign in -k is there to show that the spring force is in the opposite direction of the displacement signifying the slope. When having the springs working parallel to each other, the displacement of the two springs will become halved. When having the the two springs working in a series, the displacement will become doubled.<br><br>Page 3:<br><br></div>]]></description>
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         <pubDate>2020-02-04 05:54:59 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092077</guid>
      </item>
      <item>
         <title>Ziad Nehad</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092115</link>
         <description><![CDATA[<div>Page 1 : </div>]]></description>
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         <pubDate>2020-02-04 05:55:18 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092115</guid>
      </item>
      <item>
         <title>Ahmed Albayyat</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092121</link>
         <description><![CDATA[<div>Page 1: The applied force and the spring force have equal sizes but in opposite directions. The units for spring constant (K) is measured in N/m . K determines the elasticity of the spring. As the value of K increases the spring becomes stiffer and less stretchy.<br><br>Page 2: In page 2 we know that physicist Robert Hooke discovered that the constant of elasticity (K) and the deformation of the spring (x) are equal to the spring force (Fs) <strong>Fs=-kx. </strong>The negative sign in k  is there to show that the spring force is going in the left direction or to the opposite direction. My prediction for question 7 was correct. I stated "The displacement will be halved<br><br>Page 3: </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:55:24 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092121</guid>
      </item>
      <item>
         <title>Reem Anwar</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092129</link>
         <description><![CDATA[<div>Page 1: the Spring force and applied force are equal. the spring force and applied force are opposite in directions. the negative sign means that the force is being applied towards the negative direction. Units for k is n/m<br>when the spring is larger in size , it becomes less stretchy and displacement of the spring decreases. displacement and applied force have a direct relationship, applied force and displacement  have the same direction. <br>page 2- <br>direction of the restoring or spring force is towards the left, the negative direction. since applied force is towards the right, the spring force will be in the opposite direction. </div>]]></description>
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         <pubDate>2020-02-04 05:55:29 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092129</guid>
      </item>
      <item>
         <title>Observations:</title>
         <author>mahmo_s_121</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092155</link>
         <description><![CDATA[<div> <strong><em>PAGE 1: </em></strong><br>The sizes for applied forces are equal in <strong>magnitude</strong> but having opposite directions. The units for K is Newton per meter <strong>(N/m)</strong>. K is the <strong>''constant of elasticity'' </strong>or a particular number that describes how elasticity is stretched. So if K is high then it results in a less stretchy spring. <br><strong><br></strong><strong><em>PAGE 2:</em></strong><strong> </strong><br>In the equation<br><strong> Fs= -kx</strong>  where the negative sign beside the <strong>K </strong>in <strong>Hooke's law</strong> refers to the stretch which is in the negative direction which is opposite to the displacement. <strong>X</strong> is the elongation or the deformation of the spring. Meaning the difference in length of the spring when stretched from its unstretched length.  <br><br><br><br><br> </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:55:46 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092155</guid>
      </item>
      <item>
         <title>Jasmine F.</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092318</link>
         <description><![CDATA[<div>Page 1: <br>For applied force the sizes are the same but they have opposite directions. <br>The unit of spring constant (k) is measured in Nm and it depends on the force and displacement of the string. <br>As the string is stiffer/less stretchy K will have a higher value and as the string is less stiff/more stretchy K will have a lower value. <br>The relationship between the applied force and displacement is linear because displacement increases or decreases by the same amount each time. <br><br>page 2<br>F is known as the spring force. K is the spring constant which shows how elastic or stretchy a material is and x is the distance which shows the difference in length of the string when it's stretched or compressed. The equation is written as  F=-kx. The negative sign shows that the spring force is in the opposite direction of displacement. <br><br><br><br><br><br></div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:56:49 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092318</guid>
      </item>
      <item>
         <title>Ziad Nehad </title>
         <author>hamam_z_121</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092321</link>
         <description><![CDATA[<div>Page1: The  Spring force is a force that is directly proportional with the applied force as they have equal magnitudes they however differentiate in direction. They are equal as K the elasticity determines the spring force. K measures the spring force its written as newton per meter(N/m). The value of K  increases as the spring is more stiff and less stretchy. A relationship is seen with displacement and its applied force as when you increase the applied force the displacement will also increase .<br>Page 2: The negative sign in hooke's law (F=-kx) is a crucial sign as it doesn't affect your calculation it is there in order to show you that the spring will go to the opposite direction and return to its original place. </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:56:52 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092321</guid>
      </item>
      <item>
         <title>Hooke&#39;s Law</title>
         <author>sheha_a_121</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092390</link>
         <description><![CDATA[<div><strong> Page 1</strong><br>The direction of the applied force and spring force are equal but are opposite in direction. The spring elasticity is measured in Newtons/meter, and as it increases it requires a greater applied force to be able to stretch the spring. The applied force and displacement have a linear relationship , so as the applied force increases  the displacement increases as well.<br><strong>Page 2 <br></strong>Physicist Hooke established that most solids behave with elastic properties. Hooke's equation is<br>Fs=-ks where k stands for the constant elasticity and x stands for the spring when stretched or unstretched. The "-" sign is to indicate that the spring force "Fs" is in the opposite direction of the displacement. When the springs are working parallel to each other the displacement will be halved but when working in a series the displacement will be doubled.<br><br></div>]]></description>
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         <pubDate>2020-02-04 05:57:31 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092390</guid>
      </item>
      <item>
         <title>Page 1</title>
         <author>sheri_a_121</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092591</link>
         <description><![CDATA[<div>The applied force and the spring force are equal in magnitude but are in opposite directions. K is the spring elasticity and the greater the spring constant, the greater the force needed to stretch the spring.  Displacement and the applied force have a linear relationship, and when the force is increased, so does the displacement.<br>Page 2<br>An English Physicist, Robert Hooke, established that most solids have elastic properties. He concluded that Fs=-kx could be used to calculate the spring force. The constant k determines how elastic a solid is and the negative sign indicates that the restoring force goes in the opposite direction of the displacement. "X" represents the displacement or the difference between the spring when it is stretched from its unstretched length.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 05:59:24 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440092591</guid>
      </item>
      <item>
         <title>Ismail Jouda</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440093454</link>
         <description><![CDATA[<div>The negative sign in Hooke's Law is conceptual and isn't used in calculations, but is refers to the opposing force exerted by the spring opposing the spring's displacement.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 06:06:27 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440093454</guid>
      </item>
      <item>
         <title>Page 1</title>
         <author>salah_a_120</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440093907</link>
         <description><![CDATA[<div>The size of the applied force and the spring force are the same, however, they are opposite in direction. The constant of elasticity is measured in N/m. The spring becomes stiffer and less stretchy when the value of k increases. The applied force is directly proportional to the displacement when k is constant. Therefore, the relationship between the displacement and the applied force is linear. <br><strong><br>Page 2<br></strong>Physicist Robert Hooke has devised a relationship between the spring force(or restoring force), the elasticity of a material, and the elongation of the deformation of the spring. When written into an equation we get: Fs= -kx <br>While there is a negative sign included in the equation, it is not used in calculation. It is simply conceptual as it shows that the spring force is in the opposite direction of the displacement.<br><br><strong>Page 3<br></strong>When two springs are working in parallel, their deformation will be half the deformation of one spring working alone.<br><br>When two springs are working in series, their deformation will be double the deformation of one spring working alone.<br><br>mg=-kx<br>The slope of a spring force- displacement graph is the spring constant. Moreover, the graph is linear.<br><br>The mass of an object is equal to its spring constant multiplied by the displacement divided by the gravity.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 06:10:39 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440093907</guid>
      </item>
      <item>
         <title>Omar Eissa</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440094909</link>
         <description><![CDATA[<div>Page 1 Observations:<br>So, the applied force have the same size and are in the opposite direction. The unit for k is N/1m because it moves 1 meter and the force applied from the spring is newtons. And if k increases the spring will become more stiff and less stretchy.<br><br>Page 2 Observations:<br>Fs=-kx <br>Fs is the spring force <br>K is the spring while it's constant and it's unit is N/1m <br><br></div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 06:16:23 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440094909</guid>
      </item>
      <item>
         <title>Zeina Maged</title>
         <author></author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440098364</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 06:36:26 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440098364</guid>
      </item>
      <item>
         <title>Zarish Zafar</title>
         <author>zafar_z_120</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440550509</link>
         <description><![CDATA[<div><strong>Page 1: </strong>The size of the applied force and the restoring force are equal; however, their directions are opposite. Also, the unit of k, the spring constant, is N/m because the spring constant depends on the applied force and the displacement of the spring. Moreover, a stiffer/less stretchy  spring has a higher value of k as more energy is needed to stretch or compress the spring. The applied force and the displacement of the spring have a linear relationship because when the force increases so does the displacement of the spring. In this case, the k is kept constant and does not change. <br><br><strong>Page 2: </strong>Robert Hooke said that most solids behave (at times) with elastic properties, even very inelastic materials behave elastically under large loads. Thus, he proposed Hooke's law. Hooke's law is Fs= -kx, Fs is the restoring force and it is in units of Newton. The k is the constant of elasticity and it describes the elasticity of a spring. The x is the elongation or deformation of the spring, in other words it is the displacement of the spring. The negative sign means that the spring force is in the opposite direction of the displacement of the spring. <br><br><strong>Page 3:</strong> I have found that if I we place two springs in parallel and apply the same applied force then the displacement of the spring is reduced by a factor of 2. However, if the springs are placed in series then the displacement is doubled. This is because elongation and the spring constant are inversely related. I also learned that a graph of spring force (in newtons) versus elongation (in meters) has a slope that is equal to the value of the spring constant ( in newtons per meter) and the graph is linear. Also, we are able to find the mass of an object by using Hooke's law and Newton's second law. We first use Hooke's law to find the spring force we then use Newton's second law to find the mass of the object by dividing the spring force by the acceleration due to gravity, g=9.8 m/s^2. </div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 19:47:53 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440550509</guid>
      </item>
      <item>
         <title>Ismail Jouda</title>
         <author>jouda_i_120</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440574759</link>
         <description><![CDATA[<div><strong>Page 1<br></strong>In the first Phet simulation we were able to observe how factors ranging from applied force, spring force, to displacement contribute to Hooke's Law. According to my observation, as the applied force increase in one direction, an equal and opposite amount of restoring force is present. (x) refers to the spring elongation. Moreover, (k) is considered a spring constant that determines elasticity. The greater the value the less elastic the material is, measured in N/m. The relationship between the displacement and applied force can be expressed in a linear graph, since they're directly proportional when k is constant. <br><br><strong>Page 2</strong><br>Physicist Hooke established that most solids behave at times with elastic properties; even inelastic materials behave elastically under large loads. Furthermore, he was able to express this behavior<br>(Fs = -kx). The equation supports our previous claim in page 1 where we claimed that displacement and applied force have a linear relationship. (Fs = -kx) is similar to (y = mx), a linear graph. The negative sign in Hooke's Law is conceptual and isn't used in calculations, but is refers to the opposing force exerted by the spring opposing the spring's displacement.<br><strong><br>Page 3</strong><br>According to the 'Parallel' and 'Series' PhET simulation they don't tend to show a similar elongation and displacement. In a parallel simulation, elongation would be half the expected when using one spring. On the other hand a series simulation, the elongation would be double the expected when using one spring. As a matter of fact, series springs tend to cover greater displacement compared to parallel springs, which supports our claim.<br>In another case in 'spring-lab',<br>mg=-kx is used instead, where the acceleration due to gravity 9.8m/s^2 is put into consideration.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 20:21:23 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440574759</guid>
      </item>
      <item>
         <title>Zeina Maged</title>
         <author>elshe_z_120</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440615201</link>
         <description><![CDATA[<div><strong><em>Page 1<br></em></strong>The applied and the spring force are equal in magnitude but opposite in direction. k is the constant of elasticity. It determines how stiff or stretchy a spring is (the greater the value, the stiffer the spring, and vice-versa), and it's measured in N/m. As the applied force increases, the displacement increases when k is constant. They have a linear, directly proportional relationship. This can be seen through the formula F = -kx, which is clearly the equation of a straight line.<br><strong><em>Page 2<br></em></strong>Robert Hooke outlined an important relationship: <strong>F</strong><strong><sub>s</sub></strong><strong> = -kx<br></strong>F is the spring/restoring force, k the constant of elasticity, and x is the elongation or displacement of the spring. The (-) sign is used to show the indirect relation between the spring force and the constant of elasticity. But it can be ignored during solving.<br>- when two springs are working in parallel, the displacement will be half that of one spring, since the same force is now pulling against two springs instead of one, so it'll be as if k has been doubled.<br>- when two springs are in series, the displacement will be double that of one spring, because the same force is applied to each individual spring, and the total displacement would therefore be the sum of the elongation of each spring. <br><strong><em>Page 3<br></em></strong>When calculating the constant of elasticity, an important thing is to make sure you're using the correct units. cm must be converted to m, g to kg, and so on. We can calculate F<sub>s </sub>by multiplying the mass in kg by 9.8 (the acceleration due to gravity). Graphing the elongation and spring force values while keeping k constant confirms their direct/linear relationship.<br>Once we know the constant of elasticity, we can use that value to determine an unknown mass.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-04 21:38:14 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440615201</guid>
      </item>
      <item>
         <title>Ahmed Mohamed</title>
         <author>moham_a_121</author>
         <link>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440726878</link>
         <description><![CDATA[<div><strong>Page 1: </strong><br>The size of applied force and the spring force are at equal length. The applied force and the spring force are opposite in direction . Constant elasticity is k and having a higher k value makes the spring stiffer. The relationship of the applied force and the displacement vector when having a constant k, they are linear to each other.<br><br><strong>Page 2:</strong><br>Hooke established a law to show the correlation between Spring force Fs, the constant elasticity k, and the elongation or deformation x with <strong>Fs=-kx</strong>. The negative sign in -k is there to show that the spring force is in the opposite direction of the displacement signifying the slope. <br><strong><br>Page 3</strong>:<br>When having the springs working parallel to each other, the displacement of the two springs will become halved. When having the the two springs working in a series, the displacement will become doubled. This happens due to the elongation and the elasticity constant being inversely related. When having the spring lab experiment, in order to find the mass of the object we use <strong>mg=-kx</strong> with gravity acting as the acceleration.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-02-05 03:40:58 UTC</pubDate>
         <guid>https://padlet.com/jenbl_m_01/4gbxiueo7vlx/wish/440726878</guid>
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