# Frictional Force

1. Nov 28, 2006

### fro

A 20kg sled has final velocity of 20m/s after it ascended a 25m tall hill.

I calculated the initial velocity as 30m/s:
$$v_0 = \sqrt{{20\frac{m}{s^2}^2}+2\times10\frac{m}{s^2}\times25m$$
$$v_0 = 30\frac{m}{s}$$

Now, if 1500J of work is being done by friction, how can I find the final velocity and if the sled will be able to climb up the hill?

2. Nov 28, 2006

### KoGs

Well the frictional force is always opposing the motion. In this case the motion is up the hill, so the frictional force is going down the hill. See which force is greater. The greater one will win out and it will go in the direction of that force.

And they have already given you final velocity, no? At least you've already used it in that calculation to calculate initial velocity.

3. Nov 28, 2006

### fro

Frictional force:
$$\frac{1500J}{25m} = 60N$$

Work done at the bottom:
$$\frac{1}{2}\times20kg\times30^2\frac{m}{s^2} = 9000J$$
Force:
$$\frac{9000J}{25m} = 360N$$

Since 360N is more than the frictional force, the sled should be able to ascend. Am I right on this part?

Last edited: Nov 28, 2006
4. Nov 28, 2006

### KoGs

That just means initially it can climb the hill. But I think they want to know if it can climb the entire hill, not just a little bit of it.

5. Nov 28, 2006

### KoGs

Do they not give you an angle of the hill? Cuz technically gravity is on the side of friction too. Unless they are already including gravity in the friction.

6. Nov 28, 2006

### fro

So I add on the weight of the sled at the top (25m). Therefore,
$$60N+\frac{20kg\times10\frac{m}{s^2}\times25m}{25m} = 260N$$

360N > 260N, so the sled is able to climb up the hill?

7. Nov 28, 2006

### KoGs

Well not exactly. Have you learned conservation of energy yet? I don't think it's required to answer this problem, but it certainly helps to understand it.

Well you have your formula for kinetic energy = 1/2mv^2 right? Ok that's fine, but don't forget v is not constant. This can be seen by a lower v_f than an initial v_i. And this makes sense because there is a force opposing the motion, thus slowing it down.

8. Nov 28, 2006

### fro

Yes, we have been introduced to conservation of energy. However, I am not sure how to apply it in this particular situation. Are you suggesting that $$KE_i = KE_f + PE_f - 1500J$$?

9. Nov 28, 2006

### KoGs

Total energy is always, always the same throughout. This is what is called conservation of energy. That is, energy is conserved. You never gain enery or lose energy. And total energy is always = kinetic energy + potential energy.

What is the formula for potential energy? PE = mgh right? Well as your sled goes higher up the hill, your h is increasing. So you are gaining potential energy. But remember total energy is conserved. So in order to gain potential energy, you must lose kinetic energy. This is again seen by the lowering of the final velocity compared to the initial velocity.

And only the kinetic energy is forcing the sled to move up the hill. But it keeps changing as we move up the hill. So really the question is asking is there a point anywhere on the hill where the kinetic energy is lower than the frictional force?

10. Nov 28, 2006

### KoGs

I'm going to bed. I'll check back later if you still having difficulties.

Actually wait, I'm pretty darn sure you need to take gravity into consideration. I guess they don't give you an angle because they just want you to work in 1 dimension, so straight up and down. But definately take gravity into consideration.

11. Nov 28, 2006

### EthanB

That is exactly the equation you'll want to use.

$$KE_i$$ is determined by your mass and initial velocity.
$$KE_f$$ is determined by your mass and final velocity.
$$PE_f$$ is determined by your mass and the height to which you'll ascend.

Your calculation for initial velocity is correct. If we are examining the highest point the sled will reach, we look at a final velocity of zero. Plug and chug!

We can't simply look at forces at certain instances. Forces will be acting along the entire path, and we have to take that into account. Energy is the most simple way to do so.

On second thought, the work done by friction may not be what they gave you. Was that for the initial situation? What do they say about that? Will we have to calculate the work that would be done over a greater distance?

Last edited: Nov 28, 2006