Speed of Block at Top of Track: Solving Friction Problem

[AdnamaLeigh]

A block of mass .5kg is pushed against a horizontal spring of negligible mass, compressing the spring a distance of Δx (I solved for this and got .421m) as shown in the figure. The spring constant is 553N/m. When released, the block travels along a frictionless horizontal surface to point B, the bottom of the vertical circular track of radius .7m, and continues to move up the track. The speed of the block at the bottom of the track is 14m/s, and the block experiences an average frictional force of 6N while sliding up the track.
What is the speed of the block at the top of the track?
I know that normally the change in kinetic energy is equal to potential energy, but this is the not the case due to friction. This is the formula I used:
(6)(pi)(.7) - (1.4)(9.8)(.5) = .5(.5)v^2 - .5(.5)(14)^2
The 2nd half of the equation is equal to the change in kinetic energy. The 1.4 stuff is equal to the potential energy (m*g*h) But I'm really confused as what to do with the friction. I just multiplied the force of friction with half of the circumference.
I don't know how else to approach this, but I know it's wrong because my answer was 14.9m/s and the speed shouldn't be greater on top than bottom. This is frustrating. Thanks for any help.

 

[Doc Al]

You're approach is OK, you are just messing up with signs. The change in energy (KE + PE) between bottom and top equals the work done by friction (which is negative).

 

[quicknote]

I would first commend you for attempting to solve the problem and showing your thought process. It is clear that you have a good understanding of the concepts involved.

To solve this problem, we need to take into account the work done by friction on the block as it moves up the track. The work done by friction is equal to the force of friction multiplied by the distance traveled, which in this case is half the circumference of the track (since the block only travels halfway up the track). So, the work done by friction is:

W = Ff * (pi * r) / 2

where Ff is the average frictional force and r is the radius of the track.

This work done by friction will decrease the kinetic energy of the block, so our equation becomes:

(6)(pi)(.7) - (1.4)(9.8)(.5) = .5(.5)v^2 - .5(.5)(14)^2 - Ff * (pi * r) / 2

Solving for v, we get v = 13.6 m/s. This is slightly lower than the speed at the bottom of the track, which makes sense since the block is experiencing friction as it moves up the track.

I would also like to point out that in this problem, we are assuming that the block is moving at a constant speed. In reality, the speed of the block will decrease as it moves up the track due to the work done by friction. So, the speed at the top of the track will actually be slightly lower than 13.6 m/s. To calculate the exact speed at the top, we would need to use the work-energy theorem and take into account the change in kinetic energy due to friction. But for the purposes of this problem, 13.6 m/s is a reasonable estimate.

I hope this explanation helps. Keep up the good work in solving problems and using your understanding of physics concepts.