Consider a 2.0 kg mass on a spring on a horizontal surface. The spring has sprin
ID: 1315424 • Letter: C
Question
Consider a 2.0 kg mass on a spring on a horizontal surface. The spring has spring constant 36 N/m. The static coefficient of friction is not known and the kinetic coefficient of friction is 0.20. The spring is stretched to 20 cm beyond its equilibrium position.
a.) If the mass is initially at rest and does not move, what is the minimum value for the static friction coefficient?
b.) If the mass is given a short push so that its initial velocity is 1.0 m/s to the left, use energy considerations to determine where the mass first comes to a stop.
Explanation / Answer
Hooke's law
F=k*x
Fr=Ms*N=Ms*m*g
Ms*m*g=K*x
Ms=(K*x)/(m*g)
Ms=(36N/m*0.2m)/(2Kg*9.8m/s^2)
Ms=0.36
Wfr=Mk*N*Xf=Mk*m*g*Xf
Uf=Ui-Wfr
1/2K.(X-Xf)^2=1/2m*Vi^2+1/2K.(X)^2- Mk*N*Xf
18(0.2-xf)^2=1+0.72+3.92Xf
18Xf^2+3.28Xf-1=0
Xf= 0.16159
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