Solve Physics Problem: Block Moving Up a Vertical Track

AI Thread Summary
A block of mass 0.640 kg is pushed against a spring with a force constant of 450 N/m, compressing it a distance x, which is calculated to be 0.528 m. The block travels to the bottom of a vertical circular track with a speed of 14.0 m/s and experiences a frictional force of 7.00 N while moving up. The discussion focuses on determining the speed of the block at the top of the track, with the height at the top assumed to be 2 meters. The calculations involve applying the conservation of energy principle, but a sign error in the gravitational potential energy was noted. The predicted speed at the top of the track is calculated to be 12.9 m/s.
KD-jay
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Homework Statement


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.
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(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).

Homework Equations


Σnc=ΔKE + ΔUg + ΔUs

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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Looks like you messed up the sign of the gravitational PE in your last equation.
 
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