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Loop on a track

by KD-jay
Tags: loop, track
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KD-jay
#1
Nov25-08, 01:12 PM
P: 7
1. The problem statement, all variables and given/known data
A block of mass 0.640 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vB = 14.0 m/s, and the block experiences an average frictional force of 7.00 N while sliding up the track.


(a) What is x?
(b) What speed do you predict for the block at the top of the track?

I've gotten (a) but can not figure out (b).

2. Relevant equations
Σnc=ΔKE + ΔUg + ΔUs

3. The attempt at a solution

(a)What is x?
Σnc=ΔKE + ΔUg + ΔUs
Σnc=(1/2)*m*vf^2 - (1/2)*m*vi^2 + mghf - mghi + (1/2)*k*xf^2 - (1/2)*k*xi^2
Σnc=(1/2)*m*vf^2 - (1/2)*k*xf^2
0=(1/2)*0.640*14^2 - (1/2)*450*xf^2
x=0.528

This is the one I'm having trouble on
(b) What speed do you predict for the block at the top of the track?
I'm assuming that the work done by the friction is F*Δr = 7.00 * Π since the displacement is half the loop. I'm also assuming that the height at the top of the loop is 2 meters.

Σnc=ΔKE + ΔUg + ΔUs
Σnc=(1/2)*m*vf^2 - (1/2)*m*vi^2 + mghf - mghi + (1/2)*k*xf^2 - (1/2)*k*xi^2
Σnc=(1/2)*m*vf^2 - mghf - (1/2)*k*xf^2
-7.00*Π=(1/2)*0.640*vf^2 - 0.640*9.81*2 - (1/2)*450*(0.528)^2
vf=12.9 m/s
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Doc Al
#2
Nov25-08, 01:30 PM
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Looks like you messed up the sign of the gravitational PE in your last equation.


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