Question

A 8.4 kg crate is pulled 4.7 m up a 30 degree incline by a rope angled 15 degrees above the incline. The tension in the rope is 130 N and the crate's coefficient of kinetic friction on the incline is 0.27. 1.) What is the workd done by the Normal Force?

1.) What is the work done by the Frictional force?

ANSWER: -47.7J

2.) What is the total work done on the crate?

ANSWER: 349J

3.) If the object started at rest, what is it speed?

ANSWER: 9.12m/s

How do I get these answers?

Answer 1

Solution,

A) workdone by normal. Force is always equal to 0.

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B) normal force on the block

N = mg cos x - T sin y = 8.4* 9.8* cos 30 - 130 sin 15

N = 37.64 N

Workdone by friction

Wf = u N d = 0.27* 37.64 * 4.7= - 47.77 J

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c) net workdone

W = (130 cos y - mg sin x - uN ) d

W = ( 130 cos 15 - 8.4* 9.8 sin 30 - 0.27* 37.64) * 4.7

W = 349 J

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dl

Using work - energy theorem

W = change in KE

349 = 0.5*8.4*v^2

v = 9.12 m/s

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