A 700 g block is released from rest at height hg above a vertical spring with spring constant k =...

Question:

A 700 g block is released from rest at height hg above a vertical spring with spring constant k = 394 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 24.0 cm.

(a) How much work is done by the block on the spring?

_____J

(b) How much work is done by the spring on the block?

_____J

(c) What is the value of {eq}h_{0} {/eq}?

_____m

(d) If the block were released from height 2.00{eq}h_{0} {/eq} above the spring, what would be the maximum compression of the spring?

_____cm

Energy conservation

We know that from the law of energy conservation that energy can neither be created nor be destroyed it can only change its forms from one to another. Therefore, the total energy of the system would remain constant at every instant.

Given

Mass of the block (m) = 0.7 kg

Compression of the spring (x) = 0.24 m

Spring constant (k) = 394 N/m

(a)

Work done by the block will be stored in the spring as potential energy, therefore

{eq}\displaystyle W_{b} = 0.5kx^{2} \\ W_{b} = 0.5*394*(0.24)^{2} \\ W_{b} = 11.35 \ J {/eq}

(b)

From the Conservation of energy

{eq}\displaystyle W_{s} = W_{b} \\ W_{s} = 11.35 \ J {/eq}

(c)

Initial energy of the block {eq}\displaystyle ( E_{i} ) = mgh_{o} {/eq}

This will be converted into the spring potenial energy, therefore

{eq}\displaystyle E_{i} = W_{s} \\ mgh_{o} = 11.35 \\ 0.7*9.81*h_{o} = 11.35 \\ h_{o} = 1.65 \ m {/eq}

(d)

If the height were {eq}\displaystyle 2h_{o} {/eq}

Therefore from the energy conservation

{eq}\displaystyle mg(2h_{o}) = 0.5kx^{2} \\ 0.7*9.81*(2*1.65) = 0.5*394*x^{2} \\ x = 0.34 \ m x = 34 \ cm {/eq}