Chapter: Kinetics of particle· Engineering Mechanics

A 60 kg crate rests on a horizontal floor for which the coefficient of kinetic friction is 0.25. It is pulled by a force of 400 N acting at 25° above the horizontal. Assuming the crate does not tip over, find its velocity 5 s after starting from rest…

Q.20 A 60 kg crate rests on a horizontal floor for which the coefficient of kinetic friction is 0.25. It is pulled by a force of 400 N acting at 25° above the horizontal. Assuming the crate does not tip over, find its velocity 5 s after starting from rest, using the impulse–momentum principle.

Figure (described in words): A crate on a horizontal floor; a 400 N force acts on the crate pulling to the right at 25° above the horizontal.

Doubtly AI Asked 27 Oct 2026 AI-generated · not yet verified Practice test mu engineering Save WhatsApp Report

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Answer: \(v=21.47\) m/s.

Forces: \(W=60\times9.81=588.6\) N.

Vertical: \(N+400\sin25^\circ=588.6\Rightarrow N=588.6-169.05=419.55\) N (the upward pull reduces the normal reaction).

Friction \(F=\mu_kN=0.25\times419.55=104.89\) N, acting backwards.

Horizontal pull \(=400\cos25^\circ=362.52\) N.

Impulse–momentum (horizontal, forward positive), all forces constant:

\[mv_1+(400\cos25^\circ-F)\,t=mv_2\]

\[0+(362.52-104.89)(5)=60\,v_2\Rightarrow60v_2=1288.2\Rightarrow v_2=21.47\ \text{m/s}\]

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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A 60 kg crate rests on a horizontal floor for which the coefficient of kinetic friction is 0.25. It is pulled by a force of 400 N acting at 25° above the horizontal. Assuming the crate does not tip over, find its velocity 5 s after starting from rest…

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