Question

5. A freely falling ball of mass m = 0.5 kg passes a window 1.5 m high.

(B) If its speed at the top of the window was 2 m/s, what will its speed be at the bottom of the window?

(A) How much did the kinetic energy of the ball increase as it fell past the window?

Answer 1

Given:

$m = 0.5\:\rm kg$

$h=1.5\:\rm m$

(A) The law of conservation of energy says

$$\frac{mv_1^2}{2}+mgh=\frac{mv_2^2}{2}$$

Hence, the speed of the ball at the bottom of the window

$$v_2=\sqrt{v_1^2+2gh}=\sqrt{2^2+2*9.8*1.5}=5.8\:\rm m/s$$

(B) The change in kinetic energy of the ball

$$\Delta E_k=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}$$

$$\Delta E_k=\frac{0.5*5.8^2}{2}-\frac{0.5*2^2}{2}=7.4\:\rm J$$