# Work done by gravity and KE

1. Jul 21, 2006

### taveuni

I feel like a fool because I can't get this, but I am simply in need of some guidance re: equation usage.

A bobsled run leads down a hill as sketched in the figure above. Between points A and D, friction is negligible. Between points D and E at the end of the run, the coefficient of kinetic friction is µ = 0.4. The mass of the bobsled with drivers is 220 kg and it starts from rest at point A.
(Basically, the bobsled is at the top of hill A, which is 50m tall. It then travels to valley B (y=0) and back up to hill C (y=30m). It then proceeds down to D (y=0) where the friction starts until E.)

I need to find out the following:

-) Find the distance x beyond point D at which the bobsled will come to a halt.

I have already determined the velocity at valley B (31.32m/s) and the work done by gravity between A and C. So, the hint says "The work done by friction is the change in kinetic energy minus the work done by gravity between D and E." But how do I get the KE? I can calculate fk (ukN=863.30) and already have Work of gravity). I just need KE. And I can't figure out how to get it, correctly.

2. Jul 21, 2006

### soljaragz

I got -105600 for the change in KE between D and E. I got 122.45 as the distance beyond D when sled stops.

isn't work done by gravity between D and E zero? since the force vector is perpendicular to the displacement.

Last edited: Jul 21, 2006
3. Jul 21, 2006

### nazzard

Hello taveuni,

I couldn't find the height for point E in your post.

A(y=50m)->B(y=0m)->C(y=30m)->D(y=0m)->E(y=?)

As there is no friction between A and D, the kinetic energy & velocity of the bobsled at point D will will be the same as at point B that you've already calculated.

As there is no friction between A and D, there is no need to calculate
to solve the problem. Simply start with the bobsled at point D with no potential energy and the kinetic energy/velocity you've calculated for point B.

The bobsled comes to a halt. Therefor the change in kinetic energy will be the same as the kinetic energy at point D.

Regards,

nazzard

4. Jul 21, 2006

### soljaragz

Im implying E is also y=0 since its the end of the run

___________________
D E

5. Jul 21, 2006

### nazzard

Maybe, but I've seen some bob sled competitions on TV where the bob runs end uphill.

6. Jul 21, 2006

### soljaragz

im phys newbie so dont get mad at me but don't you have to calculate the work done by gravity between A and C?

W[g] = mgd = 220(9.8)(50-30) = 105600

so the potential energy at C as well as kinetic energy at D will be 105600.

7. Jul 21, 2006

### nazzard

Last edited: Jul 21, 2006
8. Jul 21, 2006

### nazzard

At point C the bobsled has potential energy and kinetic energy.

On a sidenote: which units do you use?

9. Jul 21, 2006

### soljaragz

oh i see. so the potential energy at C (105600 J) plus the kinetic energy at C add up to the beginning potential energy.

well then so is the answer
-110000J = 862.4x J

Last edited: Jul 21, 2006
10. Jul 21, 2006

### taveuni

I am now even more confused than I was. What about friction? If you are saying that KE at B is the same as at D, then it'll be

KE=mv^2/2=(220)(31.32^2)/2 = 107903.66 J.

Okay, that makes sense, I guess. Then we have to consider PE, which somehow you all found to be 105600 J, but I don't know how you go that. I already found Fk=863.28 J.

So what do I do with these numbers? I am so confused.

11. Jul 21, 2006

### Staff: Mentor

The sled starts out at point A with a certain mechanical energy (pure potential energy, since it starts from rest). It manages to get to point D without any energy loss, thus its KE at that point equals the original PE at point A. (Measure the PE from the y=0 level.)

So, at point D the sled has a given KE. The sled comes to rest at point E. Are you given the height of point E? If points D and E are at the same height, then all of the KE at point D is used up as "work" against friction by the time it gets to point E. (If D and E are not at the same height, you'll need to consider the change in PE between those points.)

12. Jul 21, 2006

### nazzard

Hello taveuni,

as DocAl pointed out, you still haven't answered if point E is situated at the same level as point D.

If that's the case I'd write it like that:

point A:

$$E_{pot}=mgy_A$$
$$E_{kin}=0\,J$$

with $$y_A=50\,m$$

point D:

$$E_{pot}=0\,J$$
$$E_{kin}=mgy_A$$

D->E:

$$E_{kin}=mgy_A=F\,x$$

with F being:

$$F=\mu mg$$

Now you can solve for x.

Again, this would only be the case if $y_D=y_E=0\,m$.

Regards,

nazzard

Last edited: Jul 21, 2006
13. Jul 21, 2006

### taveuni

First off, sorry I forgot - yes, the height of E is the same as D, 0.

So, I did as you said:

KE = mgy = µmgx

where KE is the kinetic energy at D, y is the height at A (50m), µ is the coefficient of kinetic friction (0.4) and x is the distance traveled to stop.

Solving for X:
mgy=µmgx

the m's and g's cancel out
so, y/µ=x

and I get-- 125m.

Right.

Now if only I could reconcile this with my class notes and find some way to understand and remember this.

Thank you so much.

14. Jul 21, 2006

### Hootenanny

Staff Emeritus
What is the problem with your class notes? Do they disagree with the above treatment or are then confusing, lacking in detail etc?