Question

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?

Short answer

The force exerted on the mower is 75 N.

The distance covered by the mower before stopping is 1.125 m.

Step-by-step solution

Step 1:Given data

• Mass of the mower = 24 kg.
• Net external force on mower = 51 N.
• Friction force= 24 N.

Determine the force exerted by the person on the mower

The friction force opposes the motion of the mower. So${\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{F}\mathbf{-}\mathit{f}$where, F_net is the net force, F is the force exerted by the person on the mower, and f is the friction force.

Substitute 51 N for F_net and 24 N for f in the above expression, and we get,

$\begin{array}{c}51\text{\hspace{0.33em}N}=F-24\text{\hspace{0.33em}N}\\ F=\left(51+24\right)\text{\hspace{0.33em}N}\\ F=75\text{\hspace{0.33em}N}\end{array}$

Hence, the force exerted on the mower is 75 N.

Determine the retardation of the mower due to the friction force

Apply Newton’s second law of motion:$\mathit{f}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{a}$where, mis the mass, and a is the acceleration.

Substitute24 kg form and24 N forf in the above expression, and we get,

$\begin{array}{c}24\text{\hspace{0.33em}N}=24\text{\hspace{0.33em}kg}\times \left(-a\right)\\ a=-\frac{24\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^{\text{2}}}{24\text{\hspace{0.33em}kg}}\\ a=-1.0{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

Determine the distance covered by the mower before stopping

Apply the law of motion:${\mathit{v}}^{\mathbf{2}}\mathbf{-}{\mathit{u}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{a}\mathit{s}$where, v is the final velocity, s is the distance covered, and u is the initial velocity.

Substitute 1.5 m/s for u, 0 for v, and -1.0 m/s² for a in the above expression, and we get,

$\begin{array}{c}0-{1.5}^2{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}=2\times \left(-1.0{\text{\hspace{0.33em}m/s}}^{\text{2}}\right)\times s\\ -2.25{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}=-2{\text{\hspace{0.33em}m/s}}^{\text{2}}\times s\\ s=\frac{-2.25{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{-2{\text{\hspace{0.33em}m/s}}^{\text{2}}}\\ s=1.125\text{\hspace{0.33em}m}\end{array}$

Hence, the distance covered by the mower before stopping is 1.125 m.