Block sliding down incline with friction on second half only

[Yashbhatt]

Homework Statement

A block starts sliding on an inclined plane inclined at an angle θ with the horizontal. The first half of the plane is frictionless and the second half has a co-efficient of friction μ. When the block reaches the bottom of the slope, it has velocity zero. Calculate the co-efficient of friction.

Homework Equations

W = mg, N = mgcosθ, F = μN

The Attempt at a Solution

Actually, my friend has solved the problem and has got the correct answer but I don't know if his method is correct. He says, if we apply a force F to an object continuously(gravitational force in this case). Then, if we apply an equal and opposite force -F, then the body will acquire a constant velocity but if we apply twice that force, the body will come to rest. So, in this case,
2 μ mg cosθ = mg sinθ and μ = 2 tanθ which is the correct value for the co-efficient of friction. As the answer is correct, the method seems correct. But I don't know for sure. Is it correct?

 

[Nathanael]

I think his explanation is excellent.



Edit:
The reason it works is only because of constant accelerations.

Since there is a constant acceleration, his method results in an equal average velocity for both halfs (and therefore equal times for which the force is being applied)

It's a sort of "symmetry" about the problem, and I think that's probably the way the problem was intended to be solved

 

[Yashbhatt]

Okay. Thanks.

 

[haruspex]

in this case,
2 μ mg cosθ = mg sinθ and μ = 2 tanθ which is the correct value

You meant μ mg cosθ = 2 mg sinθ, right?

 

[Yashbhatt]

Oh yes. Was in a hurry. So . . .
Can you show how does one go about proving it because just saying so might not be enough.

 

[haruspex]

Oh yes. Was in a hurry. So . . .
Can you show how does one go about proving it because just saying so might not be enough.

There are other ways of expressing what is essentially the same argument.
E.g. in terms of work, the downslope component of the gravity acts over the full length of the slope: W = L mg sinθ. The same work is done against friction, since no KE left at the end: (L/2) mg cos θ.