Force Exerted by a Lawnmower| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.5

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:

From the problem, we know that the net external force is 51 N. This net force is the difference between the force exerted by the man and the opposing friction force. That is

$net\:F=F-f$

So, we can solve for the force by the person to the mower.

$F=net\:F+f$

$F=51\:N\:+24\:N$

$F=75\:N$

We shall use kinematical equations to solve for the distance traveled by the mower after force F is removed. When F is removed, only f is now the only present force in the mower, so we can compute for the acceleration using the negative force.

$a=\frac{F}{m}$

$a=\frac{-24\:N}{24\:kg}$

$a=-1\:m/s^2$

We shall use this acceleration to compute for the distance travelled.

$v^2=\left(v_0\right)^2+2ax$

$\left(0\right)^2=\left(1.5\:m/s\right)^2+2\left(-1.0\:m/s^2\right)x$

$x=1.125\:m$

The mower will go 1.125 meters before stopping when the force F is already removed.