Q4.3-5PE - College Physics (Urone)

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?

Step-by-Step Solution

Answer

The force exerted on the mower is 75 N.

 

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

Step 1:Given data

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

Step 2: 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.

Step 3: 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, m is the mass, and a is the acceleration.

 

Substitute 24 kg for m and 24 N for f in the above expression, and we get,

 

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

Step 4: 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.