In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

Solution:

We can isolate the mower and expose the forces acting on it. This is the free-body diagram.

There are two forces acting on the mower in the horizontal directions:

1. \textbf{F}:This is the force exerted by the person on the mower, and it is going to the right. This is the first unknown in the problem. We treat this as a positive force since it is directed to the right. The value of this force is F=51 \ \text{N}
2. \textbf{f}: This is the friction force directed opposite the motion of the mower. We treat this as a negative force because it is directed to the left. The value of this force is f=24 \ \text{N}.

Part A. The third force in the figure is the net force, F_{net}. This is the vector sum of the forces F and f. That is

\begin{align*}
F_{net} & =F-f \\
51 \ \text{N} & = F-24 \ \text{N} \\
F & = 51 \ \text{N} +24 \ \text{N} \\
F & = 75 \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The person should exert a force of 75 N to produce a net force of 51 N.

Part B. When the force F is removed, the friction is now the only force acting on the mower. The friction is acting opposite the direction of motion. The direction of motion is indicated by the blue arrow in the figure below, and the friction force is the red arrow.

Using Newton’s Second Law, we can solve for the deceleration of the mower.

\begin{align*}
\Sigma F & = ma \\
-24 \ \text{N} & = \left( 24 \ \text{kg} \right) \ a \\
\frac{-24 \ \text{N}}{24 \ \text{kg}} & = \frac{\cancel{24 \ \text{kg}}\ \ a}{\cancel{24 \ \text{kg}}} \\
a & = \frac{-24 \ \text{N}}{24 \ \text{kg}} \\
a & = -1 \ \text{m/s}^{2} \\
\end{align*}

Using this deceleration computed above, we can solve for the distance traveled by the mower before coming to stop.

\begin{align*}
\left( v_{f} \right)^{2} & = \left( v_{o} \right)^{2}+2a\Delta x \\
\left( 0 \ \text{m/s} \right)^{2} & = \left( 1.5 \ \text{m/s} \right)^{2}+2\left( -1 \ \text{m/s}^{2} \right)\Delta x \\
 0 & = 2.25-2 \Delta x \\
2 \Delta x & = 2.25 \\
\frac{\cancel{2}\Delta x}{\cancel{2}} & =\frac{2.25}{2} \\
\Delta x & = 1.125 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}