# A large statue in a crate is being moved. The mass of the statue and the crate is 150 kg. As the statue slides down a ramp inclined at 40.0°, the curator pushes up, parallel to the ramp’s surface, so that the crate does not accelerate. If the statue slides 3 m down the ramp, and the coefficient of kinetic friction between the crate and the ramp is 0.54, calculate the work done on the crate by each of the following: (a) the gravitational force, (b) the curator, (c) the friction force, and (d) the normal force between the ramp and the crate.

So far i figured out that (a) the gravitational force and (d) the normal force do no work on the crate because they are perpendicular to the displacement. I tried to calculate the work done by the (c) friction force by using Fs=m*a=150*0 because there is no acceleration. But i am not sure if this is correct. And i do not know how to figure out the work done by (b) the curator.

Any help would be appreciated, thank you!

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1. Search

The statue of Isaac Newton is crated and moved from the Oxford Museum of Natural History and Science for cleaning

same question there

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2. Ws = M*g = 150*9.8 = 1470 N.

Fp = 1470*sin40 = 945 N.=Force parallel to the ramp.

Fn = 1470*Cos40 = 1126 N. = Normal force

Fk = u*Fn = 0.54 * 1126 = 608 N. = Force
of kinetic friction.

Fp-Fc-Fk = M*a.
945-Fc-608 = M*0 = 0.
337-Fc = 0.
Fc = 337 N. = Force of the curator.

a. Work = Fg*d = 1470 * 3 = 4410 J.

b. Work = Fc*d = 337 * 3 = 1011 J.

c. Work = Fk*d = 608 * 3 = 1824 J.

d. Work = Fn*d = 1126 * 3 = 3378 J.

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3. This is wrong

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4. **THIS IS THE RIGHT ANSWER NOT THE ONE ABOVE!**
weight = mg = 150kg*9.8m/s^2 = 1470N

Wgrav = 1470N * sin(40) * 3m = 2835 J

Fparallel = 1470N*sin(40) = 945N
Fnormal = 1470N*cos(40) = 1126N
Fkinetic = 1126N*0.54 = 608N

Wfriction = 608N * 3m * cos(180) = -1824J

Fcurrator = Fparallel - Fkinetic = 945N - 608N = 337N

Wcurrator = 337N * 3m * cos(180)= -1011J

Wnormal = 0, because the normal force is acting perpendicular to the direction of motion

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