A man is to pull a 50 .0 kg crate across a level terrain with a force of 50.0 N

Question

A man is to pull a 50 .0 kg crate across a level terrain with a force of 50.0 N. (a) If he does 4.00 x 102 J of work on the crate while exerting the force horizontally, through what distance must he have pulled it? (b) If he exerts the same force at an angle of 450 with respect to the horizontal and moves the crate through the same distance, how much work does he do on the crate? (c) What is the coefficient of kinetic friction, if the work done by friction is - 4.30 x 102J?

Explanation / Answer

a) W=F*d = 4 * 10^2/50 = 8 Metres ||||||||||||||||||||||| b) F=F/sqrt(2) 4 *= 10^2/1.414/50* = 282.88 J |||||||||||||| c)F=W/d = 4.30 * 10^2/8 = 53 ----> 53=u*500 ---> u=0.106

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