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 μk=0.47.a) How large would the coefficient of static friction μs have to be to keep the glider from springing back to the left when it stops instantaneously?

Answer 1

To keep the glider from springing back to the left when it stops instantaneously, the coefficient of static friction μs would have to be equal to or greater than the force exerted by the spring.

Step-by-step explanation:

To keep the glider from springing back to the left when it stops instantaneously, the coefficient of static friction μs would have to be equal to or greater than the force exerted by the spring. The force exerted by the spring can be calculated using Hooke's Law: F = -kx, where F is the force, k is the spring constant, and x is the displacement. In this case, x is the maximum displacement of the spring, which can be found using the equation for the potential energy stored in the spring: PE = 0.5kx^2. Rearranging the equation, x = sqrt(2PE/k).

The potential energy stored in the spring can be calculated using the kinetic energy of the glider when it stops instantaneously: KE = 0.5mv^2, where m is the mass of the glider and v is its velocity. Rearranging the equation, PE = KE = 0.5mv^2.

Substituting the values into the equations and solving for x, we get x = sqrt(2(0.5mv^2)/k).

Therefore, the coefficient of static friction μs would have to be large enough to provide a force equal to or greater than the force exerted by the spring, which can be calculated using the displacement of the spring.