Question

A 20 kg box is initially sliding across the floor at 10 m/s. If the force of friction on the box is 50 N, how far will the box slide before coming to a stop?

Answer 1

Question#19087

A 20 kg box is initially sliding across the floor at 10 m/s. If the force of friction on the box is 50 N, how far will the box slide before coming to a stop?

Solution:

Let:

$$m = 20 \, kg$$ $$v_0 = 10 \, m/s$$ $$F = 50 \, N$$

S-?

S = $v_0 t - \frac{1}{2} a t^2$, were $t$ - time to a stop

Such as $v = a t, t = \frac{v}{a}$

$$S = \frac{v_0^2}{a} - \frac{1}{2} \frac{v_0^2}{a} = \frac{1}{2} \frac{v_0^2}{a}$$

According to the second Newton's law:

$$a = \frac{F}{m}$$ $$S = \frac{1}{2} \frac{m v_0^2}{F}$$ $$S = \frac{1}{2} \frac{20 \cdot 10^2}{50} = 2 \, m$$

Answer: 2 m.