# Homework Help: Intuition about forces

1. Aug 24, 2014

### Yashbhatt

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
W = mg, N = mgcosθ, F = μN

3. 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?

2. Aug 24, 2014

### 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

Last edited: Aug 24, 2014
3. Aug 24, 2014

### Yashbhatt

Okay. Thanks.

4. Aug 24, 2014

### haruspex

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

5. Aug 24, 2014

### 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.

6. Aug 24, 2014

### haruspex

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.