How Does an Inclined Plane Affect Mechanical Energy and Work?

[bobbob]

A block of mass M is being puled up an inclined plane at a constant speed, by an attached rope that exerts a tension T. The block is pulled a distance L. The plane makes an angle theta with the horizontal, and the coefficient of kinetic friction between the block and the incline is uk.

a) Let E=PE + KE of the block. What is the change in the mechanical energy E of the block? What is the change in the potential energy of the block?

b) Verify that deltaE = Wnoncons, the change in the mechanical energy of the block is equal to the work done by the non-conservative forces acting on the block.

c) what is the Wgrav, the work done by gravity on the block? Check that Wgrav = -delta PE.

d) Combining your results find the total work done on the block, Wtot = Wcons + Wnoncons.
Verify that Wtot = delta KE

any help would be much appreciated..not sure how to do this part. thanks

 

[LowlyPion]

Welcome to PF.

Start with a) then. The block moves at constant speed, so what is the Δ in KE? It moves L, at an angle of θ, so how much higher does it get after moving L? What does that make the Δ in PE of gravity?

For the rest of it, what is the non-conservative force? Friction? And the work to friction is what? W = F*d = μ *m*g*cosθ ?

 

[nlightner]

a) The change in the mechanical energy E of the block can be calculated by subtracting the final mechanical energy from the initial mechanical energy. Since the block is being pulled at a constant speed, its kinetic energy remains unchanged. Therefore, the change in the mechanical energy is equal to the change in potential energy, given by ΔE = ΔPE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the change in height of the block (which is equal to Lsinθ).

b) The work done by non-conservative forces is given by Wnoncons = Fdcosθ, where F is the force applied (in this case, the tension T in the rope) and d is the displacement of the block (Lsinθ). Therefore, Wnoncons = TcosθLsinθ. On the other hand, the change in mechanical energy is ΔE = mgh = mgLsinθ. By equating these two expressions, we can see that ΔE = Wnoncons, as required.

c) The work done by gravity on the block is given by Wgrav = mgh. Substituting the value of h (Lsinθ), we get Wgrav = mgLsinθ. On the other hand, the change in potential energy is given by ΔPE = mgh = mgLsinθ. Therefore, Wgrav = -ΔPE, as required.

d) The total work done on the block is equal to the sum of work done by conservative and non-conservative forces. Therefore, Wtot = Wcons + Wnoncons. From part a), we know that Wcons = ΔE = mgh = mgLsinθ. From part b), we know that Wnoncons = TcosθLsinθ. Therefore, Wtot = mgLsinθ + TcosθLsinθ. This can be further simplified to Wtot = (mg + Tcosθ)Lsinθ. From part a), we also know that ΔKE = 0. Therefore, Wtot = ΔKE, as required.