# Basic mechanical energy conservation problem: block moving on an inclined plane

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Homework Statement:
A block of weight ##150N## is pulled up, by a force, on an inclined plane, reaching a height of ##2m##.
The force makes a total work of ##400J##. The block starts and ends its path at zero velocity.

How much work has the friction force made, during the lift of the block?
Relevant Equations:
none
I'm aware of the triviality of this problem, but I'm quite confused on how to solve it.
I thought that, for the conservation on energy, if the block starts from zero velocity its energy at the end of the path must also be zero: ##mgh-L_{tot}+L_{friction}=0##... but that is all but convincing me.

jbriggs444
Homework Helper
I am not clear on what you are asking. You want to know how to solve the equation you have written? Or you want reassurance that the approach you have taken and the equation that you have written are correct?

Gold Member
I am not clear on what you are asking. You want to know how to solve the equation you have written? Or you want reassurance that the approach you have taken and the equation that you have written are correct?
Yes, I am not sure about the approach I've taken to solve the situation. What I've written is based on an intuitive deduction, and I can't really get the "real physics" behind

jbriggs444
Homework Helper
Yes, I am not sure about the approach I've taken to solve the situation. What I've written is based on an intuitive deduction, and I can't really get the "real physics" behind
Yes, conservation of energy is the approach that you are expected to use. And yes, you have written a correct equation.

Personally, I'd likely have written the equation using the work-energy theorem: "change in energy = work done".

The change in energy is the change in kinetic energy plus the change in potential energy. But we know that the kinetic energy is zero before and after. So there is just potential energy to worry about. The change in potential energy is ##mgh##, of course. So that goes on the left hand side.

The work done is the total work (somebody's hand on the block or whatever) ##L_\text{tot}## minus the frictional force ##L_\text{friction}##. So that goes on the right hand side.

You end up with $$mgh=L_\text{tot}-L_\text{friction}$$Which is equivalent to what you've already written.

greg_rack
Steve4Physics
Homework Helper
Gold Member
A few points which might help…

1. It’s worth noting that we don't say a force ‘makes’, or ‘has made’ work. We say a force ‘does’ or ‘has done’ work. The phrase ‘work done’ is quite common.

2. You are using ‘L’ to represent ‘work done’. This is unusual.

3. It’s very easy to make mistakes due to signs (+/-) so I hope the following will help.

4. Work done by a force is positive if the force and displacement are in the same direction (no more than 90º difference in directions). Energy is then being supplied to the system by the force.

5. But when friction does work due to sliding, the frictional force and displacement are in opposite directions (180º). This makes the work done by the frictional force negative. Friction is removing mechanical energy from the block and turning it to thermal energy. So expect the work done by the frictional force to be negative. If it’s not, it’s wrong!

6. It is often most convenient to express values as changes (Δx), where:
Δx = (final value of x) - (initial value of x)
A positive value of Δx means x has increased, a negative value means x has decreased.

7. An importnant point is that work done by weight is minus the change in potential energy. For example: when something falls, the work done by its weight is positive (force and displacement in the same direction); but gravitational potential energy decreases, Δ(GPE)<0.

Work done by weight = -(change in GPE) = -Δ(mgh) = -mgΔh

8. Applying conservation of energy to the block, if we add the work done by every force, the total must equal the change in kinetic energy. So in the current problem we could write:

(work done by applied force) + (work done by weight) + (work done by friction) = Δ(KE)

That's the form I'd use. But if you want to use gravitational potential energy, then it's:
(work done by applied force) - Δ(GPE) + (work done by friction) = Δ(KE)

9. It is essential to get the correct signs when putting values into the above equation.

10. The '400J' given in the question is (slightly misleadingly) referred to as 'total work'. In fact it is best described as the work done by the applied force.

Lnewqban and greg_rack
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