Question

Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.
A.
2.05 m/s2
B.
2 m/s2
C.
2.1 m/s2

Answer 1

Answer:

Approximately $2.05\; {\rm m\cdot s^{-2}}$.

Explanation:

The net force on the girl would be:

$\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}$.

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be $123.0\; {\rm N}$. The girl would exert a reaction force on the rope at the same magnitude ($123.0\; {\rm N}\!$) in the opposite direction. This force would translate to a $123.0\; {\rm N}\!\!$ force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be $123.0\; {\rm N}$. Since the mass of the boy is $m(\text{boy}) = 60\; {\rm kg}$, the acceleration of the boy would be:

$\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}$.