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The statue of Isaac Newton is crated and moved from the Oxford Museum of Natural History and Science for cleaning
same question there
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!
4 answers
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.
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.
This is wrong
**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
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
1 answer