A 176g block is launched by compressing a spring of constant k =200N/m a distanc

Question

A 176g block is launched by compressing a spring of constant k=200N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction ?=0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure.(Figure 1)

After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Explanation / Answer

This is solvable from the Work Energy Theorem...

Initial Energy = .5kx2 = (.5)(200)(.15)2 = 2.25 J

When it goes through the frictional patch it will lose Energy from the Work by Friction

W = Fd = umgd = (.27)(.176)(9.8)(.85) = .396 J

Every time it goes through that patch it will lose .396 J

Start with 2.25 - .396 for one pass = 1.85 J left

Then a second pass back to the spring 1.85 - .396 = 1.46 J

Then a third pass to the curve 1.46 - .396 = 1.06 J

Then a fourth pass back to the spring 1.06 - .396 = .667 J

Then a fifth pass back to the curve .667 - .396 = .270 J

Now, it will not have enough energy to make it all the way back to the spring

Apply W = Fd = umgd

.270 = (.27)(.176)(9.8)(d)

d = .581 m

Now take .85 - .581 to measure from the left side...

That distance = .269 m (26.9 cm from the left edge)

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