# A 3-kg block rests on top of a 2-kg block supported by but not attached to a spring of constant...

## Question:

A 3-kg block rests on top of a 2-kg block supported by but not attached to a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block and (b) the maximum height reached by the 2-kg block.

## Energy Conservation:

Energy is a conserved quantity which is only converted from one form to another. For example, if an object is placed on top of a compressed spring, when the spring is released, the potential energy stored in the spring due to the compression is converted to the kinetic energy of the object. When the object reaches its highest point, all the energy has now been converted to gravitational potential energy due to the height of the object.

In our case, we have a total of 5 kg mass supported by a spring. Note that the weight of this 5 kg mass is equal to the spring force given by Hooke's law as the product of the displacement x of the spring from the equilibrium position and its spring constant k.

{eq}\begin{align*} 40 \ N/m \ \times \ x \ &= \ 5 \ kg \ \times \ 9.81 \ m/s^2\\ \\ \therefore \ x \ &= \ \dfrac{49.05 \ N}{40 \ N/m}\\ \\ &= \ 1.22625 \ m \end{align*} {/eq}

This is the extension of the spring due to the blocks. The potential energy stored in the spring is

{eq}U_s \ = \ 0.5 \ k \ x^2 \ = \ 0.5 \ \times \ 40 \ N/m \ \times \ (1.22625 \ m)^2 \ = \ 30.07378125 \ J {/eq}

a: When the 2 kg mass is released, all the spring potential energy is transfered to it as kinetic energy. Therefore, the maximum kinetic energy the 2 kg mass can have equals the spring potential energy calculated above. Let v be the maximum speed reached by the block. Therefore

{eq}\begin{align*} 0.5 \ \times \ 2 \ kg \ \times \ v^2 \ &= \ 30.07378125 \ J\\ \\ v^2 \ &= \ \dfrac{30.07378125 \ J}{1 \ kg}\\ \\ \therefore \ v \ &= \ \sqrt{30.07378125 \ m^2/s^2}\\ \\ &= \ \mathbf{5.48 \ m/s} \end{align*} {/eq}

correct to three significant figures.

b: When the block reaches its maximum height h, all the energy it has will have been converted to gravitational potential energy, with a maximum equal to the calculated potential energy of the spring.

{eq}2 \ kg \ \times \ 9.81 \ m/s^2 \ \times \ h \ = \ 30.07378125 \ J {/eq}

Solving for h, we get

{eq}h \ = \ \dfrac{30.07378125 \ J}{2 \ kg \ \times \ 9.81 \ m/s^2} \ = \ \mathbf{1.53 \ m} {/eq}

correct to three significant figures.