Question

A block of mass $\sf{m}$ is kept on the top of an inclined plane whose angle of inclination is $\sf{\theta}$ with the horizontal. The vertical height of the incline plane is $\sf{h.}$ At the end of the inclined plane, there is a spring attached. Given that, the spring constant is $\sf{k\: N/m,}$ find the maximum compression of spring.
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Kindly solve using $\mathfrak{Work\:-\:energy\:theorem,}$ not by conservation of mechanical energy.​
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Answer 1

Given:

A block of mass $\sf{m}$ is kept on the top of an inclined plane whose angle of inclination is $\sf{\theta}$ with the horizontal. The vertical height of the incline plane is h. At the end of the inclined plane, there is a spring attached. Given that, the spring constant is k N/m.

To find:

The maximum compression of the spring.

Calculation:

• Work-Energy theorem states that the total work done by all the forces is equal to the change in kinetic energy of the object.

So, final velocity of the block will be zero in this case because of the following reason:

• When the block is going down the inclined plane, initially its velocity increases and becomes √(2gh) at the end of the incline.

• However , after coming in contact with the spring, the spring force reduces the velocity of the block and its final velocity becomes zero.

Now, applying Work-Energy theorem:

$\therefore \: W_{gravity} + W_{spring} = \Delta KE$

$= > \:mgh - \dfrac{1}{2}k {x}^{2} = 0 - 0$

$= > \:mgh - \dfrac{1}{2}k {x}^{2} = 0$

$= > \: \dfrac{1}{2}k {x}^{2} =mgh$

$= > \: {x}^{2} = \dfrac{2mgh}{k}$

$= > \: x = \sqrt{ \dfrac{2mgh}{k} }$

So, final answer is:

$\boxed{ \bold{\: x = \sqrt{ \dfrac{2mgh}{k} }}}$