91影视

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

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,

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

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

Apply Newton鈥檚 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,

$\begin{array}{c}24\text{鈥凬}=24\text{鈥刱驳}脳\left(-a\right)\\ a=-\frac{24\text{鈥刱驳}鈰�{\text{m/s}}^{\text{2}}}{24\text{鈥刱驳}}\\ a=-1.0{\text{鈥刴/蝉}}^{\text{2}}\end{array}$

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,

$\begin{array}{c}0-{1.5}^2{\text{鈥刴}}^{\text{2}}{\text{/s}}^{\text{2}}=2脳\left(-1.0{\text{鈥刴/蝉}}^{\text{2}}\right)脳s\\ -2.25{\text{鈥刴}}^{\text{2}}{\text{/s}}^{\text{2}}=-2{\text{鈥刴/蝉}}^{\text{2}}脳s\\ s=\frac{-2.25{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{-2{\text{鈥刴/蝉}}^{\text{2}}}\\ s=1.125\text{鈥刴}\end{array}$

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