A 1.150 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= 0.170 m. Find the effective spring constant of the system. Submit Answer Tries 0/12 The glider is now released from rest at x= 0.170 m. Find the maximum x-acceleration of the glider. Submit Answer Tries 0/12 Find the x-coordinate of the glider at time t= 0.470T, where T is the period of the oscillation. Submit Answer Tries 0/12 Find the kinetic energy of the glider at x=0.00 m.

k = 5.29

a = 0.78m/s²

KE = 0.0765J

Explanation:

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0  

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

hypothesising or postulating

volts times amps 6 x 0.5 3w