# Homework Help: Highest velocity reached on a hill

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1. Aug 20, 2015

### Drizzy

1. The problem statement, all variables and given/known data
Alexander sled down a hill and on a snow-covered horizontal leveled ground. He wanted to calculate the highest velocity that the sled can reach. The hill is 3 meters high and 6 meters long. When Alexander sled down the hill he traveled 12,5 meters on the snow-covered ground. Alexander then makes the further assumption that the frictional force is 85% of frictional force on the ground.

Calculate the highest velocity that can be reached by the sled.

3. The attempt at a solution

I know that the highest velocity is at the bottom of the hill. Basically when he is about to touch the ground. And I am assuming that I have to calculate the acceleration on the hill and then on the ground so that I can solve the problem. I also know that I am going to use this formula: v^2 - v0^2 = 2as

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2. Aug 20, 2015

### PeroK

It's not clear what you're assuming regarding friction. Is it that the coefficient of friction is 0.85 on the ground?

3. Aug 20, 2015

### Drizzy

no! The frictional force on the hill is 85% of the frictional force that is on the ground. it doesn't say how much the frictional constant is :/

4. Aug 20, 2015

### PeroK

Okay, so why not assume the frictional forces are $F$ and $0.85F$ and see what you get?

5. Aug 20, 2015

### Drizzy

i don't know how to solve the problem. Where do I start?

6. Aug 20, 2015

### PeroK

Start at the top of the hill! You know the height, length of hill, $g$, $0.85F$. That's enough to express the velocity at the bottom of the hill in terms of these variables.

7. Aug 20, 2015

### Drizzy

okay but I need the angle to calculate the gravitational force

Cuz the sled is tilted downwards

8. Aug 20, 2015

### PeroK

If you know the height and the length, then you can work out the angle, surely?

9. Aug 20, 2015

### Drizzy

which angle do I need? the top or the bottom one? Sorry for asking so much :(

10. Aug 20, 2015

### PeroK

You don't need the angle if you think about energy. I suggest that's a better approach in any case.

11. Aug 20, 2015

### Drizzy

okay mgh but I dont have the mass of the sled

12. Aug 20, 2015

### PeroK

So what? Leave it as m.

It stands to reason that if the problem is solvable, the mass is not relevant.

13. Aug 20, 2015

### Drizzy

(g=10)
Energy = 30*m

(mv^2)/2 = 30*m
v^2 = 60
v = sqrt (60)

Is this right?

14. Aug 20, 2015

### PeroK

15. Aug 20, 2015

### Drizzy

oh I forgot...

the friction on the hill is not equal to the force cuz the velocity is not constant. the friction = normal force * mu
F(friction)=mu * F(normal)

Now I need the angle

16. Aug 20, 2015

### PeroK

You "know" the frictional force. It's 0.85F. Think energy again: work done by friction.

17. Aug 20, 2015

### Drizzy

but its not 0.85F it is 0,85 * the friction of the ground . which i dont know what to call

18. Aug 20, 2015

### PeroK

This is why you learn algebra in maths! If you always had the numbers, you'd only need a calculator. The magnitude of the force is also not relevant: only the relationship between the friction on the slope and on the ground. We'll call it $F$ on the ground, hence $0.85F$ on the slope.

Now you have two variables: $m$ and $F$.

You work though the problem using these as variables (isn't algebra great?) and look for the moment when you can cancel them out. Like you did with $m$ when you looked at motion down the slope without friction. You worked out the energy, potential and kinetic, then cancelled the $m$ to get the velocity. This is the same strategy but it's a bit harder to get to the point where you can cancel out $F$ and $m$ and solve for $v$.

This is real physics, by the way, which putting numbers into a calculator is not. Enjoy it if you can!

19. Aug 20, 2015

### Drizzy

I don't know how to use the friction in an equation. We have a force down the hill and the friction and they are opposite of each other. So we have F=F1-Ffriction

F = m * a

F1 is the force that makes the sled go downwards.

m*a=F1-0,85F

Is this right so far?

20. Aug 20, 2015

### PeroK

Yes, but think Energy. It's much simpler when you have a force and distance. How much energy is lost to friction?