[Aiyume]

A mass M is hauled from ground level up an inclined plane that makes an angle [0] with the horizontal by means of a rope passing over a frictionless pulley. The mass is pulled along until it reaches a height of H.
a)Show that if the contact between the mass and ramp is frictionless, the work done by the tension in the rope (or equivalently, by the person hauling on the rope) is independent of the angle [0].
b)Calculate the work done by the tension in the rope as a function of the ramp angle [0] if the coefficient of kinetic friction between the mass and the surface is U_k.

I have no idea how to do (a). I thought angle is needed to do the calculation.
For (b): I don't know if I did it right or not...
W=Fd
f_k = U_k N
N=mg cos [0]
F=U_k mg cos [0]

[Unknown008]

(a) Think in terms of the Principle of Conservation of Energy here.

When you move the mass from initial to final position, what energy are you giving it?

(b) It's correct so far. Now you need to find the distance the mass is moved along so that you can work out the work done, which is given by:

Work Done = Force x Distance moved in direction of force

Post what you get! :)

[Aiyume]

(a)
Tension on the inclined plane = Mg sin [0]
Distance of the inclined plane = H/ sin[0]
W = Fd = Td = (Mg sin[0]) (H/ sin[0]) = MgH
Is this correct?

[Aiyume]

(b)
f_k = u_k N
N = Mg cos [0]
F = u_k Mg cos [0]
W = Fd
d = H/sin[0]
W = (u_k Mg cos [0] H)/ (sin[0])

Is this correct?

[Unknown008]

Yes, you didn't have to go through all of this, you know.

Initially, the body is considered to have zero energy.

You give it kinetic energy to move and slowly, all the kinetic energy provided is converted to potential energy, until no more work is done, and the block becomes stationary.

So, the energy is potential energy.

EDIT: (b) Yes, this is correct. Well done! :)

[Aiyume]

Thanks!