Conservation of energy of speeding truck

[tre2k3]

I tried many ways to perfomr this problem, but i just can't figure out whether or not I am using the right values.

"A 25 ton truck loses it brakes and reaches the bottom of a hill with a speed of 60 mph.
Fortunately, there is a runaway truck ramp which is inclined at angle of 15 degrees to the horizontal."
Assuming no losses, what distance does the truck travel along the runaway truck ramp? (miles) ?

The KE of the truck just before it goes up the ramp is 6010000 ft-lb.


What do I need to do to get the distance the truck travel along the ramp? The length of the ramp is 1 mile if that means anything.

Heres a formula I've used : KE1+PE1+Work=KE2+PE2+Eloss

 

[Doc Al]

Since there are no losses, mechanical energy is conserved. Measure PE from the bottom of the ramp, then the intial KE will be completely transformed to PE. Find the height the truck must rise, then figure out how far it must have traveled up the ramp to gain that height. Pay attention to units!

 

[tre2k3]

yeah, I think the units will get me. I had to convert from ton and mph to kg and m/s and divide that by 1.36 J just to get ft-lb.

 

[tre2k3]

Since there are no losses, mechanical energy is conserved. Measure PE from the bottom of the ramp, then the intial KE will be completely transformed to PE. Find the height the truck must rise, then figure out how far it must have traveled up the ramp to gain that height. Pay attention to units!

Since I kind of new to this concept, how exactly do I go about measuring the bottom of the ramp. Do I set the value of h to 0 in mgh. Does the length of the ramp matter at all?

 

[Doc Al]

Do I set the value of h to 0 in mgh.

Yes, at the bottom of the ramp set h = 0.

Think this way: Set $1/2 m v^2 = mgh$. Once you find h, relate h to the distance up the ramp (d): $d sin\theta = h$.

 

[tre2k3]

nevermind about the length of the ramp, that was justg the distance til the truck gets to the ramp.

 

[tre2k3]

Yes, at the bottom of the ramp set h = 0.

Think this way: Set $1/2 m v^2 = mgh$. Once you find h, relate h to the distance up the ramp (d): $d sin\theta = h$.

i got it :rofl:

 

[tre2k3]

One more thing. I having trouble understandin this one.
"If instead of there being no losses, there is gravel on the runaway truck ramp that provides a constant force of 2620 pounds in the opposite direction of motion. What distance does the truck travel along the runaway truck ramp in this case?" (miles)

Do I calculate the energy loss? And do I use cos180 in the equation also?

 

[Doc Al]

Do I calculate the energy loss?

Yes, you must calculate the work done by that resistive force. Mechanical energy is no longer conserved: the change in energy will equal the work done by that resistive force. You'll find that the truck will stop in a shorter distance up the ramp.

And do I use cos180 in the equation also?

Yes, since the work done by that force is negative.

 

[tre2k3]

since i don't know the height in the equation, how do I relate that to the poisition

 

[Doc Al]

since i don't know the height in the equation, how do I relate that to the poisition

The same as before: see post #5. You can write your energy equation using only d (distance up the ramp) and not h. Then you'll solve for d, of course.