A 16.8 kg block is dragged over a rough, horizontal surface by a constant force

Question

A 16.8 kg block is dragged over a rough, horizontal surface by a constant force of 123 N acting at an angle of angle 34.9 above the

horizontal. The block is displaced 17.4 m and the coefficient of kinetic friction is 0.205. Find the work done by the 123 N force. The

acceleration of gravity is 9.8 m/s2 . Answer in units of J =1755.29

Find the magnitude of the work done by the force of friction. Answer in units of J=336.846

Find the work done by the normal force. Answer in units of J

What is the net work done on the block? Answer in units of J

Explanation / Answer

work done by force = force * displacement * cos theeta = 123 * 17.4* cos 34.9 = 1,755.289044 joules

friction force = mu * normal reaction

normal reaction = mg - F sin theeta = 16.8*9.8 - 123*sin 34.9 = 94.2660 N

so friciton = 94.2660 * 0.205 = 19.32454 N

work by friciton = friciton * distance = 19.32454 * 17.4 = 336.247 J

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