A 15.0 kg stone slides down a snow-covered hill (the figure (Figure 1)), leaving
ID: 1433081 • Letter: A
Question
A 15.0 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 2.30 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80. respectively What is the speed of the stone when it reaches point B?Explanation / Answer
A. The loss of gravitational potential energy equals the gain of kinetic energy.
Ki + Ui = Kf + Uf + W
0.5*3*v1^2 + mgh = 0.5m*v2^2 + 0
v2 = sqrt(v1^2 + 2gh)
v2 = sqrt(12^2 + 2*9.81*20) = 23.16 m/sec
B. In above equation
this time U1 = 0, U2 = 0, K2 = 0
k1 = 0.5*m*v1^2
W = Wf + Ws = uk*m*g*s + 0.5*k*x^2
0.5*m*v1^2 - uk*m*g*(100 +x) - 0.5*k*x^2 = 0
0.5*15*23.16^2 - 0.2*15*9.81*(100 + x) - 0.5*2.3*x^2 = 0
x = 20.41 m
c. When the spring is compressed x the force it exerts on the stone is
Fs = kx = 2.3*20.41 = 46.94 N
fs = us*m*g = 0.8*15*9.81 = 117.72 N
The spring force is less than the maximum possible static friction force so the stone remains at rest.
Let me know if you have any doubt.
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