A 15.0 kg block is dragged over a rough, horizontal surface by a70.0 N force acting at 20.0 degree angle above the horizontal. The block is displaced

asked 2021-02-21
A 15.0 kg block is dragged over a rough, horizontal surface by a70.0 N force acting at 20.0 degree angle above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3. Find the work done on the block by ; a) the 70.0 N force,b) the normal force, and c) the gravitational force. d) what is the increase in the internal energy of the block-surface system due to friction? e) find the total change in the kinetic energy of the block.

Expert Answers (2)

1) thework done by the 70 N force is
\(\displaystyle{W}_{{1}}={F}{d}{\cos{\theta}}={70}\cdot{5}{\cos{{20}}}={328.89}\ {J}\)
2) The normal force is perpendicular to the displacement ofthe motion, so it does no work.
3) The gravitational force is perpendicular to the displacement, or the displacement in the vertical direction is the work done is 0
Best answer


\(W_F = Fd = 70 \cdot  5 \cos 20 = 142.82 J\)


\(N = mg \cos x = 15 \cdot 9.8 \cos 90 = 0 J\)


\(W_g = -mg \cos 90 = 0 J\)


\(W_f = -u d (mg-F \sin x) = -0.3 \cdot 5 \cdot ( 15 \cdot 9.8 - 71.9 \sin 20)\)

\(W_f = -183.61 N\)


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