Answer to Question #260294 in Field Theory for Pranali

Question

A 366.0 g block is dropped onto a vertical spring with a spring constant k = 225.0 N/m. The block becomes attached to the spring, and the spring compresses 0.230 m before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?

What work is done by the spring?

What was the speed of the block just before it hit the spring?

Answer 1

(a) We can find the work done by the block's weight as follows:

$$W_g=mgx=0.366\ kg\times9.8\ \dfrac{m}{s^2}\times0.23\ m=0.82\ J.$$

(b) We can find the work done by the spring as follows:

$$W_{s}=-\dfrac{1}{2}kx^2=-\dfrac{1}{2}\times225\ \dfrac{N}{m}\times(0.23\ m)^2=-5.95\ J.$$

(c) We can find the speed of the block just before it hit the spring from the law of conservation of energy:

$$KE_f-KE_i=W_g+W_s,$$ $$0-\dfrac{1}{2}mv_0^2=W_g+W_s,$$ $$v_0=\sqrt{\dfrac{-2(W_g+W_s)}{m}},$$ $$v_0=\sqrt{\dfrac{-2\times(0.82\ J-5.95\ J)}{0.366\ kg}}=5.3\ \dfrac{m}{s}.$$