# I have a question related to work done against gravity on an incline, frictionless plane that I'm hoping you can help me with. Here is the question: "A 10kg box is pushed up a frictionless...

I have a question related to work done against gravity on an incline, frictionless plane that I'm hoping you can help me with.

Here is the question:

"A 10kg box is pushed up a frictionless incline 5.0m long. The surface is inclined at 30 relative to the horizontal. Determine the work done against gravity to push the box at a constant speed all the way up the incline.(Hint: Gravity is a vertical acceleration so your displacement must also be vertical). **5 marks.**"

I have two possible answers, with reasons for each. The vertical displacement of the box is 2.5m [(5m)sin(30°)] so this would mean that the work done against gravity is W=mgh=(10kg)(9.81m/s²)(2.5m)=245.25 J.

I am confused, because you are not simply lifting the box, you're pushing it up an incline, so shouldn't this require less effort?

If this is the case, my answer would be

Parallel Force=Fgsinθ=(98.1N)sin(30°)=49.05N

W=Fdcosθ=(49.05N)(5m)cos(30°)=212.393 J. This would make sense to me because you are exerting less energy to raise the box the same height, which is the basis of levers and ramps. Which of these answers is correct and why? Am I making this more complicated than needed by involving power in some way?

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### 2 Answers

Also, note that it is important to understand what kind of work you are calculating:

The problem is asking to calculate the work done against gravity. It is implied that some external force is pushing the force up the incline, but we do not no anything about this force. All we know that the work performed by this force must be equal and opposite to the work performed by gravity, because the box is moving with constant speed (if the external work was greater, the box would accelerate, if it was smaller, the box would not move.)

To calculate the work performed by gravity, we need to multiply the magnitude of the force (mg) by the magnitude of displacement (L = 5 m) by the cosine of angle between these two angles (120 degrees, since gravity is pointing down, and displacement vector is pointing up the incline.)

So `W_(gravity) = mgLcos(120) = -245 J`

Notice that the negative sign is consistent with the fact that the gravity is working against the motion, which is up the incline.

So the work performed by external force must be `W_(external) = -W_(gravity) = 245` J.

Another way to look at the formula is that it is the product of gravity, mg, and "vertical displacement", which is Lsin(30) = Lcos(60), which is how your first result was obtained. And, as mentioned in the answer above, this work equals the change in the box's potential energy.

The problem with your second answer is that if you assume that the external force is parallel to the incline and balances out the component of the gravity parallel to incline (which is okay to assume, since this results in the box's motion with constant speed), and calculate this force, as you have done correctly

`F_(||) = mgsin(30) = 49 N`

then to calculate the work of this force, you need to multiply it by the magnitude of the displacement, L = 5m, (not vertical displacement !) and the cosine of the angle between the `vecF_(||)` and the displacement, which is 0, because both parallel force and displacement are directed up the incline. Since cos(0) = 1,

`W_(||) = mgsin(30)L = 245 J` .

The work of external force is 245 J, same result we obtained before.

Your first answer is correct. It's a common misconception that simple machines (levers, ramps, pulleys, wheel and axle, wedges, screws) make less work. They do not make less work. The reason simple machines make it feel like you are doing less work is because they usually reduce *force. * Each machine allows a user to reduce force by trading distance. A ramp will reduce how much force a person uses, but the box must move further.

For example, I can lift a box straight up to a height of 2 meters. If it requires a force of 10 Newtons, then the total work being done is 20 J (force x distance). Or I could push it up to that same height using a ramp. The box reaches the same height, and I didn't have to push/pull so hard. Suppose my force was reduced to 2 Newtons using the ramp. That's great, but that means I had to push the box up a 10 meter long ramp.

Your question is equivalent to using no ramp at all, because it is frictionless. No work is being done to overcome the friction of the ramp, so it's no different than a dead lift. Your horizontal and vertical *force* may have been reduced, but the work done to get the box to that height is the same with or without the ramp.

Another way to look at it is to use gravitational potential energy. It's measured in Joules the same as work. If 50 J of work was used to lift a box to a certain height, then it has 50 J of gravitational potential energy. Think about your ramp. If there was no ramp, a friction ramp, or a no friction ramp, the box still ends up at the same *height. *In all cases it has the same amount of potential energy. Same potential energy, same amount of work.

A frictionless ramp is the only kind of ramp that will give you ideal work (work without friction). Any other ramp actually requires more *work* to lift the box, because some of your input energy is being used to overcome the friction of the ramp itself.

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Thank you so much for your detailed answer and the clarification! I was obviously confusing work and force. Thanks again! :-)