Work Energy Power

JEE Main level, numerical answer, medium

81 of 183

Question 81 of 183

A block of mass 'mm' (as shown in figure) moving with kinetic energy EE compresses a spring through a distance 25cm25cm when, its speed is halved. The value of spring constant of used spring will be nENm−1nEN{m}^{-1} for n=n=_____.

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Answer: 24

Using work −- energy theorem
Wnet=∆KE{W}_{net}=∆KE
Wnet=(Kf−Ki){W}_{net}=\left({K}_{f}-{K}_{i}\right)
⇒−12Kx2=12m(v2)2−12mv2⇒−12Kx2=E4−E\Rightarrow -\dfrac{1}{2}K{x}^{2}=\dfrac{1}{2}m{\left(\dfrac{v}{2}\right)}^{2}-\dfrac{1}{2}m{v}^{2} \Rightarrow -\dfrac{1}{2}K{x}^{2}=\dfrac{E}{4}-E
⇒12Kx2=3E4⇒K=3E2x2\Rightarrow \dfrac{1}{2}K{x}^{2}=\dfrac{3E}{4} \Rightarrow K=\dfrac{3E}{2{x}^{2}}
⇒K=3E2×(14)2=24ENm−1\Rightarrow K=\dfrac{3E}{2\times {\left(\dfrac{1}{4}\right)}^{2}}=24EN{m}^{-1}
Thus, the value of n=24n=24.

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