What Is the Speed of a Block at Different Points on a Frictionless Hill?

[misyg]

A 250g block slides down a frictionless hill. If the hill is 1.2m high and the speed of the block is 5 m/s when it is halfway down, what was the speed of the block at the bottom and at the top?

I tried using the equation KE(top)+GPE(top)+SPE(top)=KE(middle)+GPE(middle)+SPE(middle) and the same for the middle+bottom But I just could not figure out the correct answer.

 

[MacDougal87]

Assuming your SPE is spring potential energy, and you're setting that value to zero, you're on the right track. Can you post your actual calculation steps? I should be able to point you in the right direction if you show me how you plugged your numbers in for your calculations =)

 

[misyg]

I set them equal to zero, the SPE, and came up with the equation for the top and middle of the hill...top(1/2mv2)+(mgh)=(1/2mv2)+(mgh)middle...(1.25v2)+(2.94)=(3.125)+(1.47)
v2=1.324
vtop=1.1507

does that sound correct.

 

[MacDougal87]

Well, the approach was fine, however it looks like you made a little mistake on converting your mass, judging by your coefficient in front of v². it looks like you used the correct mass for the rest of your values however, so it must have been just a little slip up for that one. Try solving it using .125 instead and see if that gets you your answer =)

 

[rwisz]

I would approach this problem by first setting up a free-body diagram to understand the forces acting on the block. Since the hill is frictionless, the only forces acting on the block are its weight and normal force from the hill.

Using the equation for the conservation of energy, we can calculate the speed of the block at the bottom of the hill. We know that at the top of the hill, the block has only potential energy and at the bottom, it has both kinetic and potential energy. Therefore, we can write the equation as follows:

KE(bottom) + PE(bottom) = KE(top) + PE(top)

Since the block has no initial velocity at the top, we can set KE(top) = 0. We also know that PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the hill. Substituting these values, we get:

KE(bottom) + 0 = 0 + mgh

Solving for KE(bottom), we get KE(bottom) = mgh

Next, we can use the equation for kinetic energy, KE = 1/2mv^2, to calculate the speed at the bottom of the hill. Substituting the known values, we get:

KE(bottom) = 1/2(0.25)(v^2)

Rearranging, we get v = √(2gh)

Substituting the values of g = 9.8 m/s^2 and h = 1.2 m, we get v = 3.43 m/s. Therefore, the speed of the block at the bottom of the hill is 3.43 m/s.

To find the speed at the top of the hill, we can use the equation for conservation of energy again. However, this time, we will set KE(middle) = 0 since the block is at rest at the middle of the hill. This will give us the following equation:

KE(top) + PE(top) = 0 + 0

Solving for KE(top), we get KE(top) = -PE(top)

Substituting the values, we get KE(top) = -(0.25)(9.8)(0.6) = -1.47 J

Using the equation for kinetic energy, we can solve for the velocity at the top of the