A block is attached to a horizontal spring of stiffness k. The other end of the spring is attached to a fixed wall. The entire system lie on a horizontal surface and the spring is in natural state. The natural length of the spring is $ℓ_{0}$ If the block is slowly lifted up vertically to a height $\frac{5}{12} ℓ_{0}$ from its initial position, which of the following is not correct:

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The work done by the $\frac{5}{12} {mgℓ}_{0} + \frac{{kℓ}_{0}^{2}}{288}$ lifting force

The work done by the spring force $- \frac{{kℓ}_{0}^{2}}{288}$

The sum of works done by all the forces on the block is zero

The work done by the gravity $\frac{5}{12} ℓ_{0} mg$

answer is A.

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Detailed Solution

Change in kinetic energy of the block is zero. Hence by work energy theorem ,option $(4) i s w r i t e .$

Again elongation produced in the spring = $\frac{l_{0}}{12} .$ So work done by the spring = - $(E l a s t i c P E s t o r e d i n t h e s p r i n g) = - \frac{k {l_{0}}^{2}}{288}$

Work done by the gravity force= - $\frac{5 m g l_{0}}{12}$

By work energy theorem , $W_{F} + W_{g} + W_{S} = 0 \Rightarrow W_{F} = \frac{5 m g l_{0}}{12} + \frac{k {l_{0}}^{2}}{288}$

Hence option $(1)$ is wrong

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