A 0.450 kg air-track glider is attached to each end of the track by two coil spr

Question

A 0.450 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.090 m.

(Image url below)

https://s1.lite.msu.edu/res/msu/physicslib/msuphysicslib/26_Oscillations1_SHM/graphics/prob33a_1031glidspr.gif


Find the effective spring constant of the system?

The glider is now released from rest at x= 0.090 m. Find the maximum x-acceleration of the glider?

Find the x-coordinate of the glider at time t= 0.450T, where T is the period of the oscillation?

Find the kinetic energy of the glider at x=0.00 m?

Explanation / Answer

Fspring = -kx so substitute in known values to get k Total energy of the system = 1/2 m v^2 + 1/2 k x^2 at x = .310m 1/2 m v^2 is zero so only potential energy. So plug in the numbers into 1/2 k x^2 to get total energy. At x=0 m potential energy goes to zero so the kinetic energy is just the 1/2 k x^2 you got before. (1/2 k x^2 at x = .310) = (1/2 m v^2 at x = 0)

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