Question

A block of mass $m$ is slowly lowered from a point where it just touches a vertical fixed spring of stiffness $k$ , till it remains stationary after the applied force is withdrawn. Find the work done by the external agent

$(a)$ in compressing the spring by a distance $x$ and

$(b)$ bringing the block to its stable equilibrium position.

Question Image

$(a)$ in compressing the spring by a distance $x$ and

$(b)$ bringing the block to its stable equilibrium position.

Accepted Answer

$(a)$ As the block is moving down, the spring pushes it up with force $F_{sp} = k x$ .

The work done by the spring is $W_{sp} = - \frac{1}{2} k x^{2}$ .

The work done by gravity is $W_{gr} = m g x$ .

The block does not change its kinetic energy as it is lowered slowly, $∆ K = 0$ .

Assuming $W_{ext}$ as the work done by the applied external force.

Applying work-energy theorem, i.e., work done by total forces is equal to change in kinetic energy of the system.

$W_{total} = ∆ K$

$W_{ext} + W_{sp} + W_{gr} = ∆ K$

Substituting $W_{sp} = - \frac{1}{2} k x^{2}$ , $W_{gr} = m g x$ and $∆ K = 0$ .

We have,

$W_{ext} - \frac{1}{2} k x^{2} + m g x = 0$ .

$(b)$ When the block remains at rest under the action of gravity and spring force, the continuously increasing spring force $k x$ will nullify the gravity force $m g$ , so we have $k x = m g$ .

This gives $x = \frac{m g}{k}$ which corresponds to the stable equilibrium position.

Now substituting $x = \frac{m g}{k}$ in the expression, $W_{ext} = \frac{1}{2} k x^{2} - m g x$ , we have,

$W_{ext} = \frac{1}{2} k {(\frac{m g}{k})}^{2} - m g (\frac{m g}{k})$

$W_{ext} = - \frac{m^{2} g^{2}}{2 k}$ .

B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 30: ILLUSTRATION 8.29

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 30: ILLUSTRATION 8.29

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