Recent content by Angela_vaal

I'm confused.

Is the collision inelastic then?

Homework Statement A 0.038-kg pet lab mouse sits on a 0.35-kg air-track cart, as shown in (Figure 1) . The cart is at rest, as is a second cart with a mass of 0.25 kg. The lab mouse now jumps to the second cart. After the jump, the 0.35-kg cart has a speed of v1=0.86m/s. What is the speed v2...

I figured it out. Thanks for all your help!

how do I know the value of the total energy?

In that case, there is a normal force. So the forces on the block are the Normal force, MG and 250cos16, 250 sin16

I would use total energy=PE+KE ?

Homework Statement A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block? Homework Equations Fk=μkN F=MA The Attempt at a...

wouldn't the potential energy increase and kinetic energy decrease? the total energy should remain conserved.

I'm not sure :(

I just reviewed it in my textbook and I was so wrong with the units. Going back to part B I got, PE: (.5)(637N/m)(.01m)2=.03185J I still don't understand how the velocity changes with each compression. would I solve for a new velocity and use the ΔKE to find the new velocity? ΔKE=...

I thought x represented distance, so I assumed it was the 4.15 cm in the beginning of the problem. I thought K was the value that was given for the compression.

Can you please explain further...

isn't that the compression of 1cm?

I'm just confused about it all! It looks like I am pluggining in the wrong numbers!