Question

1. a person pulls horizontally on a rope attached to a 50 kg box that is sliding away from them. the force of friction between the box and the floor is 125 n. if they wish to slow the box at a rate of 4 m/s², how hard must they pull on the rope?

Explanation:

Step1: Apply Newton's second law

Newton's second law is \(F_{net}=ma\). The net force \(F_{net}\) is the sum of the frictional force \(F_f\) and the tension force \(T\) (since they act in the same direction to slow the box). So \(F_{net}=F_f + T\).

Step2: Substitute values into the formula

We know \(m = 50\space kg\), \(a=- 4\space m/s^{2}\) (negative because it's decelerating), and \(F_f = 125\space N\). Substituting into \(F_{net}=ma\) gives \(F_f+T=ma\).
Then \(T=ma - F_f\).
Substitute \(m = 50\space kg\), \(a=-4\space m/s^{2}\), \(F_f = 125\space N\) into the equation:
\(T=(50\times(- 4))-125\)
\(T=-200 - 125\)
\(T=-325\space N\). The negative sign indicates the direction of the force (opposite to the motion of the box, which is consistent with the problem's requirement of slowing the box). The magnitude of the force is \(325\space N\).

Answer:

The person must pull with a force of \(325\space N\).