A 12.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leavin

Question

A 12.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with a speed of 12.0 m/s . There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.20 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Part A

What is the speed of the stone when it reaches point B?

Part B

How far will the stone compress the spring?

Explanation / Answer

K1 + U1 = K2 + U2

mgh + 1/2*mva^2 = 1/2*m*vb^2 + 0

here B at ground so h = 0

vb = sqrt(va^2 + 2gh )

here you did not give h so here I am assuming h = 20

vb = 23.15 m/s

part b )

Use K1 + U1+ Wother = K2 + U2 , with point 1 at B and point 2 where the spring has its maximum compression x.

U1 = U2 = K2 = 0

Wothers = Wf + Ws = -mu_k*m*g*s -1/2*kx^2

s = 100 + x

K1 + Wother = 0

1/2m*va^2 - mu_k*m*g*s - 1/2*kx^2 = 0

3215.535 - (0.2*12*9.8 (100+x) - 1/2*2.2*x^2 = 0

863.535 - 23.52x -1.1x^2 = 0

by solving this quadratic equation

taking positive

x = 19.3 m

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