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!

**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

This is wrong

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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

Well, it seems like you're in a bit of a pickle, but don't worry, I'm here to help you out with some clownish knowledge!

Let's break it down:

(a) The gravitational force: Since the crate is sliding down the ramp, the gravitational force is doing negative work because it is acting in the opposite direction of the displacement. The work done by the gravitational force can be calculated using the formula: work = force × displacement × cos(theta). In this case, the force is the weight of the crate, which is given by mass × gravity, where mass = 150 kg and gravity = 9.8 m/s². The displacement is 3 m, and theta (the angle between the force and displacement) is 40°. Plug in the values and calculate the work done by the gravitational force.

(b) The curator: The curator is pushing up, parallel to the ramp's surface, so the work done by the curator is in the same direction as the displacement. The force the curator exerts is equal in magnitude but opposite in direction to the gravitational force. Therefore, the work done by the curator can be calculated using the same formula as (a) but with the force exerted by the curator instead.

(c) The friction force: The work done by the friction force can be calculated using the formula: work = force × displacement. The force of friction can be found using the equation: force of friction = coefficient of friction × normal force. Since there is no motion perpendicular to the ramp's surface, the normal force is equal in magnitude but opposite in direction to the gravitational force. Calculate the normal force and use it to calculate the force of friction. Then plug in the values and calculate the work done by the friction force.

(d) The normal force: As you said, the normal force does no work on the crate because it is perpendicular to the displacement. Therefore, the work done by the normal force is zero.

I hope this helps you calculate the work done by each force involved! Remember, even in the face of a physics problem, a little laughter can lighten the load. Keep smiling, my friend!

To calculate the work done on the crate by each force, there are a few steps you need to follow. Let's go through them one by one:

(a) Gravitational force:
The work done by the gravitational force can be calculated using the formula: work = force * distance * cos(theta), where force is the weight of the crate and statue, distance is the displacement down the ramp, and theta is the angle between the force and the displacement.

In this case, the gravitational force is acting vertically downward, which is perpendicular to the displacement down the ramp. Therefore, the angle between the force and displacement, theta, is 90 degrees. Since cos(90) = 0, the work done by the gravitational force is zero.

(b) Curator:
The curator is pushing up on the crate parallel to the ramp's surface, so the force applied by the curator is also parallel to the displacement down the ramp. When a force is parallel to the displacement, the work done is given by: work = force * distance.

To calculate the force exerted by the curator, we need to consider the forces acting on the crate in the vertical direction. The gravitational force downwards is balanced by the normal force upwards, since the crate is not accelerating vertically. Therefore, the force applied by the curator is equal in magnitude to the weight of the crate and statue, which is 150 kg * 9.8 m/s^2 = 1470 N.

Now, using the distance the crate slides down the ramp (3 m), substitute the values into the work formula: work = 1470 N * 3 m = 4410 J.

(c) Friction force:
The work done by the friction force can be calculated using the formula: work = force * distance * cos(theta), where force is the frictional force, distance is the distance traveled down the ramp, and theta is the angle between the force and the displacement.

The frictional force can be calculated using the formula: friction force = coefficient of friction * normal force. In this case, the normal force is equal in magnitude to the weight of the crate and statue, which is 150 kg * 9.8 m/s^2 = 1470 N. Therefore, the frictional force is 0.54 * 1470 N = 793.8 N.

Since the friction force acts opposite to the direction of motion, the angle theta between the force and the displacement is 180 degrees. Thus, cos(180) = -1. Substituting the values into the work formula: work = 793.8 N * 3 m * (-1) = -2381.4 J (negative sign indicates work is done against the motion).

(d) Normal force:
The normal force does not do any work on the crate because it is perpendicular to the displacement down the ramp.

To summarize:
(a) Work done by the gravitational force = 0 J
(b) Work done by the curator = 4410 J
(c) Work done by the friction force = -2381.4 J
(d) Work done by the normal force = 0 J

Therefore, the work done on the crate by (a) the gravitational force is zero, (b) the curator is 4410 J, (c) the friction force is -2381.4 J, and (d) the normal force is zero.

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.