Work done against friction for a car with 125 KJ KE.

[Molly1235]

"A 1200kg car as kinetic energy 125kJ at the bottom of a 20 degree slope. It rises to a height of 10m. Calculate work done against friction."

Relevant equations:

Work = force x distance in direction of that force

Work = KE = 1/2 x m x v^2

I'm not really sure where to start...tried several things but haven't got anywhere.

For example:

Work done = KE
125000J = 1/2 x 1200 x v^2
V^2 = 125000/600 = 208.3
V = 14.4 m/s

Or

GPE at top of slope = mgh
1200 x 9.81 x 10 = 117,720

Energy lost = 125000 - 117720 = 7280 J

But neither of these answer the question.

Also tried

Distance traveled = 10/sin20 = 29.24m

So f = 125000/29.24 = 4274.9J buy apparently that's wrong.

Someone please help? :-)

 

[Doc Al]

GPE at top of slope = mgh
1200 x 9.81 x 10 = 117,720

Energy lost = 125000 - 117720 = 7280 J

This is the one you want.

If that's not right: Are you sure it says "rises to a height " of 10m? Maybe they meant that it travels up the ramp a distance of 10 m?

 

[Molly1235]

This is the one you want.

If that's not right: Are you sure it says "rises to a height " of 10m? Maybe they meant that it travels up the ramp a distance of 10 m?

I tried it that way as well and none of the answers match up (it's multiple choice). Hm, perhaps I copied the answers/some of the question down wrong!

 

[Molly1235]

This is the one you want.

If that's not right: Are you sure it says "rises to a height " of 10m? Maybe they meant that it travels up the ramp a distance of 10 m?

So just to confirm, would the energy lost be the same as the work done against friction?

 

[Doc Al]

So just to confirm, would the energy lost be the same as the work done against friction?

Yes, assuming friction is the only source of energy "loss".

 

[nasu]

I tried it that way as well and none of the answers match up (it's multiple choice). Hm, perhaps I copied the answers/some of the question down wrong!

What are the choices?