An air-track glider of mass 0.109 kg is attached to the end of a horizontal air

Question

An air-track glider of mass 0.109 kg is attached to the end of a horizontal air track by a spring with force constant 19.5 N/m

(a) With the air track turned off, the glider travels 8.0 cm before it stops instantaneously. How large would the coefficient of static friction s have to be to keep the glider from springing back to the left?

(b) If the coefficient of static friction between the glider and the track is s= 0.55, what is the maximum initial speed v1 that the glider can be given and remain at rest after it stops instantaneously? With the air track turned off, the coefficient of kinetic friction is k= 0.45. m/s

Explanation / Answer

a) Sum of the horizontal forces on the block (opposite friction is positive) = 0 <---- no movement
0 = F - friction
F = friction

From Hooke's Law
F = k*x

Sum of the vertical force on the block (up is positive) = 0
0 = N - m*g
N = m*g

Definition of friction
friction = u*N
friction = u*m*g

Plug back in and solve for u
u*m*g = k*x
u = (k*x) / (m*g)

Plug in numbers
u = (15.0 N/m * 0.083 m) / (0.101 kg * 9.81 m/s^2) = 1.26

b) I'm getting stuck on this part...
Use Hooke's law to get the intial spring elongation. Then you have to use energy balance and work to find the maximum velocity. I'll keep working on it for a few more minutes, but I'm not getting the same answer.

xf = u*m*g / k = 0.64 * 0.101 kg * 9.81 m/s^2 / 15 N/m = 0.0423 m

When I back it to the relaxed spot, I find the velocity is 0.328 m/s

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