Question

(a)

A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,300 N/m, and compresses the spring 0.250 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.

 (b)

 What If? The spring now expands, forcing the block back to the left. Does the black reach point B?

 If the block does reach point B, how far up the curved portion of the track does it reach, and if it does not, how far short of point B does the block come to a stop? (Enter your answer in m.)

Answer 1

Velocity of block at point B

0.5 m vb^2 = mgh

vb^2 = 2 gh

vb^2 = 2* 9.8* 3

vb = 7.668 m/s

velocity at point C

0.5 m vc^2 = 0.5 k x^2

15* vc^2 = 2300* 0.25^2

vc = 3.1 m/s

Using conservation of energy

Workdone by friciton = change in kE

u mg d = 0.5 m ( vb^2-vc^2)

u * 9.8* 6 = 0.5* (7.668^2 - 3.1^2)

u = 0.418

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Velocity at B during moving to the left

u g d = 0.5* ( vc^2 - vb'^2)

0.418* 9.8* 6 = 0.5* ( 3.1^2 - vb'^2)

vb'^2 = - 39.55

Since Vb' is negative it means the block wocan't clear the frictional length.

So it will stop at

0.418* 9.8* d' = 0.5* vc^2

d' = 1.173 m

It would short form B by

D = 6 - d' = 4.827 m

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