Box sliding down frictionless incline

AI Thread Summary
A box sliding down a frictionless ramp accelerates due to gravitational potential energy converting to kinetic energy. Initially at rest, the box reaches a speed of 4 m/s at the bottom. When starting with an initial speed of 3 m/s, conservation of energy principles are applied to determine the final speed. The energy equation combines both gravitational potential and kinetic energy, leading to a final speed of approximately 4.9 m/s. The discussion emphasizes the importance of energy conservation in solving the problem.
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Homework Statement



A box initially at rest at the top of a frictionless ramp is allowed to slide to the bottom. At the bottom its speed is 4 m/s. Next, the box is again slid down the ramp, but this time it does not start from rest. It has an initial speed of 3 m/s at the top. How fast is it going when it gets to the bottom?

Homework Equations



No idea...


The Attempt at a Solution



Since the length of the ramp, mass of the block, nor the time it takes to get from the top of the ramp to the bottom is given... I don't know where to start with this one.
 
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EDIT: What I had done was wrong :).

Matterwave is correct, just use conservation of energy.
 
Last edited:
I'm not sure I understand your equation...
vf meaning velocity and ?
vi is initial velocity, correct?
 
I'm not sure I understand your equation...
vf meaning velocity and ?
vi is initial velocity, correct?

Yes, I'm just using a subscript to denote the initial and final states of the velocity.
 
Perhaps Jegues' method is simpler, but I would approach this in terms of energy conservation. Potential energy at top = kinetic energy at bottom. The m's will cancel out.
 
so a=(vf-vi)/t

a=4/t

a=(vf-3)/t

set them equal to each other
4/t = vf-3/t

t's cancel out and we get 4= vf-3
and vf=7 :)
 
Bump, I edited my post above. As Matterwave suggested conservation of energy is what you're looking for.

EDIT:
so a=(vf-vi)/t

a=4/t

a=(vf-3)/t

set them equal to each other
4/t = vf-3/t

t's cancel out and we get 4= vf-3
and vf=7 :)

This is what I had initially suggested. It is wrong. Use conservation of energy, sorry in advance for any confusion.
 
KE= 1/2mv^2
GPE= mgh

and that would give us h= .81716
but how would I use that solve for vf is vi=3?
 
You generated the height using the first case good!

Now for second case, in the intial state what types of energy are there? In the final state what types of energy are there?
 
  • #10
At the top of the ramp, all energy is in Gravitational Potential (mgh) and KE.
and at the bottom there is only KE.

1/2vi^2 + gh = 1/2vf^2
12.499 = 1/2vf^2
and vf = 4.9
 
  • #11
Looks correct to me!
 
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