Kinetic Energy/ Friction

chaoseverlasting · Mar 19, 2007

1. The problem statement, all variables and given/known data
There are two fixed inclined planes as shown (ABC and DGEF). AB=DE=y, BC=EF=x. The coeff of friction between the wedges and the mass m is [tex]\mu[/tex].

A small mass M is released at the point A and reaches the bottom with a speed V1. The same block is released from point D and reaches F with speed V2. Which is the correct option:

V1>V2
V1<V2
V1=V2.

3. The attempt at a solution

I thought since the potential energy is same here in both cases, but the path length DGF is greater than AC, therefore more energy must be dissipated by friction and so V1>V2.

But the answer is V1=V2.

How?

Doc Al · Mar 19, 2007

While the path lengths are different, so are the frictional forces, which depend on the angle. Figure out the work dissipated in each case.

chaoseverlasting · Mar 21, 2007

But the angles arent given

Doc Al · Mar 21, 2007

chaoseverlasting said: ↑

But the angles arent given

You don't need the angles--everything you need can be expressed in terms of X & Y.

chaoseverlasting said: ↑

1. The problem statement, all variables and given/known data
There are two fixed inclined planes as shown (ABC and DGEF). AB=DE=y, BC=EF=x.

I just realized that this description does not match your diagram. In the diagram, you have both triangles with the same height (that would make the problem too easy!). Per your description, the second triangle must be shorter.

What's the correct description of the problem?

Log in or Sign up

Kinetic Energy/ Friction

chaoseverlasting

Attached Files:

untitled.bmp

Doc Al

Staff: Mentor

chaoseverlasting

Doc Al

Staff: Mentor

Coefficient of kinetic friction and Energy (Replies: 1)

Friction and Kinetic Energy (Replies: 1)

Friction and kinetic energy (Replies: 3)

Kinetic energy dissipation by friction (Replies: 5)

Kinetic energy and friction force? (Replies: 9)