- #1
AdnamaLeigh
- 42
- 0
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.
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.
