Calculating Work on an Inclined Plane: A 15N Force

[ritwik06]

Homework Statement

A force of 15 N is required to pull up a body of mass 2 kg through a distance of 5m along an inclined plane making an angle of 30 degree with the horizontal. (If you need, I can give the figure with the question.) Calculate:
1)the work done by the force in pulling the body
2)the work done against the force due to gravity (g=9.8 N/kg)
3)account for the difference in the magnitude of the above two readings.

Homework Equations

Work=Force*Displacement*cosine of the angle

The Attempt at a Solution

Part 1) Work done =15*5=75 J as the force and displacement are in the same direction, i.e., 30 degree with the horizontal)

Part 2)Work done against gravity=mgh=2*9.8*(5*cos30)=49 J
5*cos 30, give the vertical displacement.

Part 3) Here lies the problem, I think that the difference is because, in first case we consider vertical displacement and the horizontal one, I mean both the components, so the work done is more. In the second case, only the vertical component counts, both for the force and the displacement.

My book gives a vague answer to this,

The difference in work=(75-49)=16 J is utilised in doing work against friction between the body and the inclined plane.

Please help me with part 3. I will provide the figure given with this question if it is needed but you. But please do try to help.

 

[Hootenanny]

You may wish to reconsider your answer to part (2). Are you sure you want the consine of the angle...?

 

[Doc Al]

Part 1) Work done =15*5=75 J as the force and displacement are in the same direction, i.e., 30 degree with the horizontal)

Good.

Part 2)Work done against gravity=mgh=2*9.8*(5*cos30)=49 J
5*cos 30, give the vertical displacement.

I assume you mean sin30, not cos30.

Part 3) Here lies the problem, I think that the difference is because, in first case we consider vertical displacement and the horizontal one, I mean both the components, so the work done is more. In the second case, only the vertical component counts, both for the force and the displacement.

My book gives a vague answer to this,

I don't quite understand your answer. But consider that the problem states "A force of 15 N is required...". To me, that means that the 15 N just allows you to move the body up the incline at constant speed. If there were no friction, what force would be needed? What must be the friction?

 

[ritwik06]

Good.


I assume you mean sin30, not cos30.


I don't quite understand your answer. But consider that the problem states "A force of 15 N is required...". To me, that means that the 15 N just allows you to move the body up the incline at constant speed. If there were no friction, what force would be needed? What must be the friction?

Oh yes I meant sin30. I am sorry for that.

I meant that while carrying the body up the inclined plane, the body is making actually two displacements, S cos30 and S sin30.
Naturally (F cos30)(S cos30)+ (F sin30) (S sin30)-(F sin30)(S sin30) = 16joules. What do you think?
F=15 N and S=5metres.

 

[Doc Al]

I meant that while carrying the body up the inclined plane, the body is making actually two displacements, S cos30 and S sin30.

OK, nothing wrong with that.

Naturally (F cos30)(S cos30)+ (F sin30) (S sin30)-(F sin30)(S sin30) = 16joules. What do you think?

Realize that your first two terms are equivalent to just F*S:
(F\cos\theta)(S\cos\theta) + (F\sin\theta)(S\sin\theta) = FS (\sin^2\theta + \cos^2\theta) = FS

So with the first two terms, you just took the long way around to calculate the work done by the applied force.

I don't understand your third term. Did you mean to write (mg)(S sin30), which is the work done against gravity?

If so, all you've done is calculate the difference. They want you to explain the difference!

 

[ritwik06]

OK, nothing wrong with that.

Realize that your first two terms are equivalent to just F*S:
(F\cos\theta)(S\cos\theta) + (F\sin\theta)(S\sin\theta) = FS (\sin^2\theta + \cos^2\theta) = FS

So with the first two terms, you just took the long way around to calculate the work done by the applied force.

I don't understand your third term. Did you mean to write (mg)(S sin30), which is the work done against gravity?

If so, all you've done is calculate the difference. They want you to explain the difference!

Exactly I did that. I mean that if you lift a load vertically (without any machine). The work done will be the product of force applied in the upward direction and the vertical distance covered. Assume that there is no diplacement in the horizontal plane.

Now I use an inclined plane. As in practice we don't use an inclined plane making an agle of 90 degree with the horizontal. Do we? So it means that the body to be lifted to some height always makes displacement in the horizontal plane as well, don't you think so? A force is required to make this displacement, right? Then, of course, work is done in moving the body in the horizontal plane, which is not (useful) required.

Thus the overall work done is more than the work done in just lifting the load vertically upwards (without any displacement in the horizontal plane). Thus this accounts for the diffrerence. Now what do you say?

I really don't get how they can say that was friction alone.

 

[ritwik06]

one more thing.
if I want to lift a load of 2 kg mass. Why should there be a force of 'mg' acting vertically upwards to lift it? "mg" is the force with which the body is pulled towards the Centre of Gravity of earth. If we apply a force in the opposite direction, the body would be in equilibrium. Now to lift it upwards, don't you think, that the force applied should be more than just "mg"?

This is another doubt which arises in my mind. But please solve my previous problem.

 

[Doc Al]

Exactly I did that. I mean that if you lift a load vertically (without any machine). The work done will be the product of force applied in the upward direction and the vertical distance covered. Assume that there is no diplacement in the horizontal plane.

OK.

Now I use an inclined plane. As in practice we don't use an inclined plane making an agle of 90 degree with the horizontal. Do we?

Right! And why is that? Hint: It's not because the work against gravity is any different using an incline.

So it means that the body to be lifted to some height always makes displacement in the horizontal plane as well, don't you think so?

Of course. If you move something up an incline, it moves horizontally as well as vertically.

A force is required to make this displacement, right? Then, of course, work is done in moving the body in the horizontal plane, which is not (useful) required.

No additional work is required to move something up an incline--unless friction is present.

Thus the overall work done is more than the work done in just lifting the load vertically upwards (without any displacement in the horizontal plane). Thus this accounts for the diffrerence. Now what do you say?

I say you are wrong. I had asked earlier: If there were no friction, what force would be needed to push the object up the incline (at constant speed)? Figure that out and then calculate the work done in that case.

one more thing.
if I want to lift a load of 2 kg mass. Why should there be a force of 'mg' acting vertically upwards to lift it? "mg" is the force with which the body is pulled towards the Centre of Gravity of earth. If we apply a force in the opposite direction, the body would be in equilibrium. Now to lift it upwards, don't you think, that the force applied should be more than just "mg"?

Sure, to start the object moving (from rest) you have to apply some net force on it (which gives it some kinetic energy). But once it's moving, it only requires an applied force of "mg" upwards to keep it going. The required work against gravity is the same.

 

[ritwik06]

I say you are wrong. I had asked earlier: If there were no friction, what force would be needed to push the object up the incline (at constant speed)? Figure that out and then calculate the work done in that case.

Is the force required = W sin30? Work done in that case = W sin 30*S

 

[Doc Al]

Is the force required = W sin30? Work done in that case = W sin 30*S

Exactly right. That's the work done against gravity. Notice that it depends on the change in height (h = sin 30*S), not on the horizontal distance covered.