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Ask QuestionPosted by Ayush Vishwakarma?? 6 years, 4 months ago
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Posted by Pawan Sidhu 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Consider a particle having mass m moving with a velocity v so, that its kinetic energy, K.E = {tex}\frac{1}{2}{/tex}mv2 and momentum, p = mv.
Thus, {tex}\mathrm{K.E}=\frac{p^{2}}{2 m} \text { or } p=\sqrt{2 m K .E}{/tex}
when, kinetic energy is increased by 300%, then new kinetic energy is given by:
K.E' = K.E + 300% of E =KE + 3KE = 4K.E
New momentum p' = {tex}\sqrt{2 m \mathrm{K.E}^{\prime}}=\sqrt{2 m \times 4 \mathrm{K.E}}{/tex} {tex}=2 \sqrt{2 m \mathrm{K.E}}=2 p{/tex}
Therefore, the Percentage increase in momentum {tex}=\frac{p^{\prime}-p}{p} \times 100=\frac{2 p-p}{p} \times 100=100 \%{/tex}
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#Aditi~ Angel???? 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
A variable force is a force whose magnitude or direction or both vary during the displacement of a body on which it acts or a force whose direction or magnitude or both change with time.
The graphical representation is given as:
Consider a small displacement element {tex}\Delta s{/tex} under force F, which is represented by a strip KLMN. As the displacement {tex}\Delta{/tex}s is extremely small, hence force F for entire strip KLMN may be taken as constant.
Therefore, Work done during elementary expansion will be equal to
{tex}\Delta W =F\Delta x{/tex}= area of strip KLMN ....................................(A)
Total work done by the variable force can be calculated by dividing the whole path into such elementary parts and in each case the work done will be equal to area of small shaded strips like KLMN. Therefore, total work done for a given displacement is given by:
W = {tex}\sum \Delta W=\sum{/tex} area of various strips = total area under F-s graph .......................................(B)
Thus, the work done by a variable force is given by the area under F-s curve.
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#Aditi~ Angel???? 6 years, 4 months ago
3Thank You