# Homework Help: Momentum and Energy of a car of mass

1. Jun 18, 2006

### MD2000

I had a couple homework questions that I was having trouble solving..I was hoping someone could help me out..

1. A car of mass 1430 kg traveling at 26 m/s is at the foot of a hill that rises 120 m in 3.8 km. At the top of the hill, the speed of the car is 10 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses.

Heres what I got..
P = W/T
W = 1.2mfvf^2 - 1/2mivi^2
W = 1/2(1430)(10)^2 - 1/2(1430)(26)^

For the T I did:
1/2(V0+V)t = X

P = [1/2(1430)(10)^2 - 1/2(1430)(26)^2] / [X/1/2(V0+V)]
P = -1950.2

Apparently this is wrong..

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2. A roller coaster reaches the top of the steepest hill with a speed of 5.4 km/h. It then descends the hill, which is at an average angle of 45o and is 40 m long. What will its speed be when it reaches the bottom? Assume µk = 0.16.

I figured out FN = mgcos(45)
(ukxmgcos45)(dxcos180) = 1/2vf^2 - 1/2(1.5)^2

And I got 9.298..am I on the right track?

2. Jun 18, 2006

### Hootenanny

Staff Emeritus
For question one you can calculate the time period knowing that the acceleration is;

$$-9.81\cdot\frac{120}{3800}$$
(Resolving the acceleration due to gravity)

Inital kinetic energy + potential energy = Final Kinetic energy + Work done against friction

You were correct with your calculation of the normal force.

3. Jun 18, 2006

### MD2000

Hmm..why wouldn't X/1/2(V0+V) work in the first scenario since we have the initial and final velocity? Also, I'm kind of confused as to how you got the time..because according to what you have posted I'm getting a huuuuge amount..1.33x10^6..

For the second equation..how were u able to plug in "work done against friction..(ukxmgcos45)(dxcos180)" in to the equation? Was it (KEf+PEf) - (KEi+PEi) = Work..and you just rearranged?

4. Jun 18, 2006

### Staff: Mentor

There's nothing wrong with using this to find the time. Where you went wrong is in finding the change in energy: you counted KE, but not PE.

5. Jun 18, 2006

### MD2000

The thing is Im still getting different numbers for both situations..and completely different answers..I jus dont understand why..

The second one I think I got..21.62?

6. Jun 18, 2006

### MD2000

There is actually another similiar problem that I'm having trouble with:

A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 11.2 m up along this slope before coming to rest. What was the average coefficient of friction?

I was planning on using

Fkdcos(thet) = (KEi+PEi) - (KEf+PEf)

Does that sound like the right approach

7. Jun 19, 2006

### Hootenanny

Staff Emeritus
Did you use the acceleration I gave you above to calculate the time period? You should have something like;

$$\text{power} = \frac{mgh - \frac{1}{2}mv_{f}^{2} - \frac{1}{2}mv_{i}^{2}}{t}$$

Note that you divide the whole change in energy (kinetic and potential) by the time taken. Perhaps if you show your working we can point out where you have gone wrong.
Looks good to me.
This also looks good to me. In this case, you must take the base of the slope to be the point of zero potential, so the PEi term drops out.

Last edited: Jun 19, 2006
8. Jun 19, 2006

### Staff: Mentor

That should be the change in total mechanical energy over time, not just the change in kinetic energy over time. Potential energy counts too.

9. Jun 19, 2006

### Hootenanny

Staff Emeritus
Thanks Doc

10. Jun 19, 2006

### MD2000

I keep on getting different numbers for T..

By using Hootenanny's acceleration method..

I used x = vit + 1/2at^2 and solved for t..and ended up getting 93.61

I then used X = 1/2(Vi + V)t and got 211.1

*scratches head*..shouldn't they give you the same thing?

11. Jun 19, 2006

### Hootenanny

Staff Emeritus
When I calculated the acceleration, I was assuming the the 3.8km is the length of the hill, i.e. the hypotenues of the triangle, is this correct?

12. Jun 19, 2006

### MD2000

Yea..3800 m is going to be the hypo..am I jus doin the math wrong?

As for the following prob I got .364..for the coefficient of friction..how does that look?

A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 11.2 m up along this slope before coming to rest. What was the average coefficient of friction?

13. Jun 19, 2006

### Hootenanny

Staff Emeritus
I'm scratching my head to, I wander why my method isn't working . I would go with your method though. As for your skier, you are correct.

14. Jun 19, 2006

### Staff: Mentor

Your calculation of the acceleration assumed that the only force acting on the car is gravity (and the normal force, of course). Not so.

15. Jun 19, 2006

### Hootenanny

Staff Emeritus
I was thinking along those lines, but I can't see the other force(s). Could you show me the light please Doc

16. Jun 19, 2006

### Staff: Mentor

I shall shine forth the light....

If the car were merely sliding without friction up the incline, then the only active force would be gravity and your value for acceleration would be correct. But, thankfully, the motor turns the tires, the tires push the ground, and the ground pushes back. It's the friction of the ground pushing against the tires that you are missing. (That's how the car is able to transform chemical energy into mechanical energy.)

17. Jun 19, 2006

### MD2000

lol. it turns out the method i was using worked out alright. thanks for the help guys. this stuff is finally making some sense..hehe

18. Jun 20, 2006

### Hootenanny

Staff Emeritus
And he was illuminated... Thanks Doc