Question

The kinetic friction force between a 64.0-kg object and a horizontal surface is 48.0 N. If the initial speed of the object is 25.0 m/s, and friction is the only force acting on the object, what distance will it slide before coming to a stop? Answer: _________ m/s (round to the nearest tenth)

Answer 1

First, use Newton's Second Law of Motion to find the acceleration of the object, considering that only the friction is acting on the object with a mass of 64.0kg:

$\begin{gathered} \Sigma F=ma \\ \\ \Rightarrow f=ma \\ \\ \Rightarrow a=\frac{f}{m}=\frac{48.0N}{64.0kg}=0.75\frac{m}{s^2} \\ \\ \therefore a=0.75\frac{m}{s^2} \end{gathered}$

Next, remember the following formula that relates the initial and final velocities of a uniformly accelerated particle with the distance that it travels and the acceleration:

$d=\frac{v_f^2-v_0^2}{2a}$

Since the object starts at 25.0 m/s and stops, then v_f=0. Since the object is decelerating, the acceleration is negative.

Replace v_0=25.0m/s, a=-0.75m/s^2 and v_f=0 to find the distance that it will slide before coming to a stop:

$\begin{gathered} d=-\frac{v_0^2}{2a}=-\frac{(25.0\frac{m}{s})^2}{2(-0.75\frac{m}{s^2})}=416.666....m \\ \\ \therefore d\approx416.7m \end{gathered}$

Therefore, to the nearest tenth, the distance that the object slided before coming to a stop is 416.7m.