Question

An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m. Initially the spring is unstreched and the glider is moving at 1.50 m/s to the right. With the air track turned off, the coefficient of kinetic friction is ?k=0.47. It can be shown that with the air track turned off, the glider travels 8.6 cm before it stops instantaneously.

Part A) How large would the coefficient of static friction ?s have to be to keep the glider from springing back to the left when it stops instantaneously?

Part B) If the coefficient of static friction between the glider and the track is ?s = 0.75, what is the maximum initial speed v1 that the glider can be given and still remain at rest after it stops instantaneously?

Answer 1

We begin with

KE = ½mv² = ½ * 0.100kg * (1.50m/s)² = 0.1125 J

a) While the spring is being stretched, that is converted into friction work and PE:

0.1125 J = µk*m*g*x + ½kx²

0.1125 J = 0.47*0.100kg*9.8m/s²*x + ½*20.0N/m*x²

This is quadratic with a negative root (irrelevant)

and a positive root at x = 0.0855 m

and furthermore we need the static friction force to equal the spring force:

µs*m*g = k*x

µs * 0.100kg * 9.8m/s² = 20.0N/m * 0.0855m

µs = 1.74 ?

b) Now work it backwards:

0.75 * 0.100kg * 9.8m/s² = 20.0N/m * x

x = 0.03675 m

KE = 0.47*0.100kg*9.8m/s²*0.03675m + ½*20.0N/m*(0.03675m)² = 0.03043 J

0.03043 J = ½ * 0.100kg * v²

v = 0.780 m/s