Question

The kinetic friction force between a 61.0-kg object and a horizontal surface is 42.0 N. If the initial speed of the object is 23.0 m/s, and friction is the only force acting on the object, what distance will it slide before coming to a stop? Answer: __________ m (round to the nearest whole number)

Answer 1

First, use Newton's Second Law of Motion to find the acceleration of the object, given that its mass is 61.0kg and the force acting on it is 42.0N:

$\begin{gathered} f=ma \\ \\ \Rightarrow a=\frac{f}{m}=\frac{42.0N}{61.0kg}=0.6885...\frac{m}{s^2} \end{gathered}$

Next, remember that the distance d traveled while an object stops from an initial speed v under an acceleration a is given by:

$d=\frac{v^2}{2a}$

Replace v=23.0m/s and the value of the acceleration found above:

$d=\frac{(23.0\frac{m}{s})^2}{2(0.6885...\frac{m}{s^2})}=384.15...m\approx384m$

Therefore, the distance that it slides before coming to a stop is approximately 384m.