A mower's dolly is used to transport a refrigerator up a ramp into a house. The refrigerator has a mass...

Question

A mower's dolly is used to transport a refrigerator up a ramp into a house. The refrigerator has a mass of 115 kg. The ramp is 2.10 meters long and rises 0.850 meters. The mover pulls the dolly with a force of 496 N up the ramp. The dolly and ramp constitute a machine.

a. What work does the work do?
b. What is the work done on the refrigerator by the machine?
c. What is the efficiency of the machine?

Please help me with this question.
I'm having trouble understanding it.

Answer 1

a. What work does the work do?
I can tell you don't understand the question. This question makes no sense.

c. What is the efficiency of the machine?
Insufficient information is given to answer this question.

Answer 2

I'm assuming your HW assignment actually asks this...

a.) What work does the MOVER do?
b.) What is the work done on the refrigerator by the machine?
c.) What is the efficiency of the machine?

Anyway.... Ive got a similar HW assignment myself, and I'm not one-hundred percent sure about my response, but I'm pretty sure its correct.

a.) Work = Force x Distance

In this case, we are told that the worker applies a force of 496 N up the ramp. This is important. The force applied is parallel to the the length of the hypotenuse of this ramp (2.10 meters). So:

Work = 496 N x 2.10m
Work = 1041.6 N x m

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b.) Work = Force x Distance

In this case, we are being asked how much work the machine (the ramp, dolly, the wheels on the dolly, etc) is doing on the refrigerator.

The first confusing factor may be the Force value. We certainly cannot directly compute what the ramp and dolly are doing on the refrigerator! However, we CAN calculate what the refrigerator is doing on the ramp and dolly.

The refrigerator itself is really only applying the force of its WEIGHT on this machine. We can see this because if the worker here stopped moving the dolly, the refrigerator would not launch itself out of the dolly at the same speed and direction it was going. Granted, it may move forward on the dolly or wobble slightly, but in this problem that is considered negligible. It is clear that the majority of the force is being applied STRAIGHT DOWN on the ramp. Therefore, weight-force should be the force value used in the work equation.

F=mg

The refrigerator has a mass of 115kg and g = 9.8 m/s^2.

F= 115kg x 9.8 m/s^2
F = 1127 N

Therefore, the refrigerator is exerting 1127 N of force on the machine. However, the question asks how much work the machine does on the refrigerator! In this case, the weight force (directed toward the ground) and the normal force (directed upwards towards the refrigerator) are equal.

We can see that this is true by observing that if the worker stops moving the dolly, that the refrigerator stays still. It will not move up into the air, and it will not move down (aka, through the ramp). Therefore, the machine (ramp, etc) MUST be applying a force equal to the weight force. At this point, it is clear that the force value of 1127 N is the right force value to plug into the work equation. So:

F = 1127N

d = ?

Now about distance... Since the force we calculated is weight force (the force applied STRAIGHT DOWN), than the distance traveled will be a vertical increase. We are given this vertical increase in distance. The ramp is .850 meters tall. Relatively, the refrigerator is traveling up this distance throughout the event. So;

d = .850 m

We now have our F and d values!

F = 1127N
d = .850 m

Work = F x d

Work = 1127N x .850 m
Work = 957.95 N x m

So the FINAL ANSWER TO PART B is 957.95 N x m

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c.) Efficiency = Output Work / Input Work

In this case the worker has "put in" 1041.6 N x m.
The output that the machine produced for us was 957.95 N x m.

Efficiency = (957.95N x m) / (1041.6 N x m)
Efficiency = .91969086...

Efficiency is approx 92%. Note that the units cancel, and that efficiency is therefore a ratio.

I hope that my explanation helps you understand both this problem and how to figure out which formulas and data to use in other problems. Peace Out!

C2H5SH

P.S. - You can google that chemical compound later : )