Can Someone Help me?

Question

A 181 g block is launched by compressing a spring of constant k=200N/m at 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 a coefficient of friction μ=0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise. After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Answer 1

You are being asked to measure the final result from the left of

the frictional surface, but there is no indication whether

the spring is on the left or right of the frictional zone.

I will assume that it is on the left.

If I am wrong, correcting the final answer should be easy.

Let

m = 181 g = 0.181 kg be the mass of the block,

k = 200 N/m be the spring constant,

c = 15 cm = 0.15 m be the initial spring compression,

d = 85 cm = 0.85 m be the length of the stretch with friction,

μ = 0.27 be the coefficient of friction,

g = 9.81 m/s² be gravitational acceleration.

The initial energy of the spring gets converted to the work of friction.

If there is any energy left after passing the distance d then the block will move up the incline.

The initial energy of the spring is

kc²/2 = 200×0.15²/2 = 2.25 J

The weight of the block is

mg

The force of friction is

mgμ

If the block goes the entire distance d, then friction will do work

mgμd = 0.181×9.81×0.27×0.85 = 0.4 J

Therefore after going the whole distance d, there is leftover energy.

This leftover energy will propel the block up the curved rise.

Once it reaches as high as it can, it will slide back again.

While going up and down the curved rise it does not lose any energy,

because the surface is frictionless.

Once it hits the frictional zone again, it will slide back losing another 0.4 J of energy.

With the remaining energy it will compress the spring (though less than the initial distance c).

That will then propel it to go on the same journey again.

This going back and and forth will continue until all the energy will get spent on the frictional surface.

The initial 2.25 J of energy allows it to go across the frictional surface

five time and have 0.25 J left for one final partial trip on the frictional surface.

Since the number of full trips across the frictional surface is odd,

the final partial trip will start on the right, and will travel some distance x.

This distance x is such that the work of friction over x will burn

the remaining 0.25 J:

mgμx = 0.25

x = 0.25/mgμ = 0.25/0.181×9.81×0.27 = 0.52 m

When measured from the left the block will stop at distance

0.85 - 0.52 = 0.33 m

Caution: My numbers involve some rounding.

This rounding gets multiplied by 5 -- the number of trips over the frictional surface.

I do not know what accuracy you are to achieve, but you might need more accuracy than I am showing.