# Work and forces

I have these 3 problems, and it doesnt tell me whether my answer is right or wrong right away. I choose the answer I thought was appropriate, and explained why. Can anyone take a look at the explanations and see if they're true? I'm sure some may be flawed.

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PhanthomJay
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That's a lot of questions in a single post! You should post each one separately with your relevant equations and attempt at a solution. For the first one, what are the forces acting on the mass in each case that do Work? Then use the definition of Work to get your answer. The one you checked is wrong.

Sorry I didnt want to spam the forum with 3 different posts all relatively the same.

So for the free fall case, forces acting on mass that do work are:
-its weight, mg

Incline:
Since there is no friction, only force that act on its movement are mg. But I'm not really sure here: the normal force acts on it as well but 'canceled out' by the w=mg, since both of those added should equal zero, no?
So the mg in this case is weaker than the free fall, since its not full gravity.

Last case:
Forces acting on the ball are mg as well as Tension.

These kind of questions usually have the answer of 'same', which it probably is in this case, but it's not my first logical guess.

I still dont see how my answer is wrong. Isnt the force of gravity strongest in the free fall? For incline, it would be only sin/cos of gravity, so only a portion, right?

PhanthomJay
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Sorry I didnt want to spam the forum with 3 different posts all relatively the same.
I understand, but when you pose several questions at once, helpers and you can get bogged down just responding back and forth to one of them, and before you know it, the post either gets several pages long, or, more likely, your other questions get lost in the abyss (black hole).
So for the free fall case, forces acting on mass that do work are:
-its weight, mg
correct; how much worK?
Incline:
Since there is no friction, only force that act on its movement are mg. But I'm not really sure here: the normal force acts on it as well but 'canceled out' by the w=mg, since both of those added should equal zero, no?
No. The normal force acts perpendicular to the incline, so it doesn't cancel out the weight. But the normal force does no work...W =Fdcostheta, but theta is 90 degrees for the angle between the force and displacement, so the work done by the normal force is 0. What's the work done by gravity?
So the mg in this case is weaker than the free fall, since its not full gravity.
mg is always full gravity. The component of the gravity force doen the plane is less than mg, but the distance travelled is longer. Again, what's the work done by gravity?
Last case:
Forces acting on the ball are mg as well as Tension.
again the tension always acts perpendicular to the motion, so it does no work.
These kind of questions usually have the answer of 'same', which it probably is in this case, but it's not my first logical guess.

I still dont see how my answer is wrong. Isnt the force of gravity strongest in the free fall? For incline, it would be only sin/cos of gravity, so only a portion, right?
See above response

PhanthomJay
Homework Helper
Gold Member
For question 2, there is no displacement relative to the truck,but plenty relative to the ground..and also a friction force acting that keeps the block accelerating at the same rate of the truck.

When you say whats the work done by gravity, are you asking for a value? Well work is force x displacement (integral, kind of) so work done by gravity for the free fall is 9.8*m*h -is that what you're asking for?

Work done by incline is: m*h*9.8cosx

For question 2, there is no displacement relative to the truck,but plenty relative to the ground..and also a friction force acting that keeps the block accelerating at the same rate of the truck.
Ahh ok..so the block IS accelerating, even though its not really moving itself?

PhanthomJay
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And regarding the third question, 'total' work and 'net' work are the same term. Since the net force is 0, the net work is___??___?

PhanthomJay
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Gold Member
When you say whats the work done by gravity, are you asking for a value? Well work is force x displacement (integral, kind of) so work done by gravity for the free fall is 9.8*m*h -is that what you're asking for?
yes, correct, now compare that to the net work done in the other 2 cases
Work done by incline is: m*h*9.8cosx
The work done by the normal force is 0. The work done by gravity is mgsintheta times the length (h/sin theta) of the incline. Or, the work done by gravity (or any conservative force) is always just the negative of the change in potential energy