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 0.700 N to displace the glider to a new equilibrium position, x= 0.270 m.

Find the effective spring constant of the system.
2.59 N/m

You are correct. Your receipt no. is 160-5399

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The glider is now released from rest at x= 0.270 m. Find the maximum x-acceleration of the glider.
1.56 m/s^2

You are correct. Your receipt no. is 160-6175

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Find the x-coordinate of the glider at time t= 0.470T, where T is the period of the oscillation.

Calculate the angular velocity and the period, then the time at which you must find the position.

Incorrect.

Tries 3/12

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Find the kinetic energy of the glider at x=0.00 m.
9.45×10^-2 J

Answer 1

Given mass-m-0.450kg Horizontal applied force-F-0.7 N Displacement of new equilibrium position-x-0.27m The restoring force is, The effective spring constant is, 0.7 0.27 2.59 N/m Time-t-0.470T The angular frequency is, 2π 2.59 0.45 -2.399 rad/s The maximum acceleration is, (2.399) (0.27-停)(04707)) = 1.55 m/s x=Acos (or = ( 0.27)cos( )(0.477) -0.27 m Kinetic energy at x-0 m The speed is, = (0.27)(2.399) = 0.647 m/s The kinetic energy is, (0.450)(0.647) = = 9.42 x 10-2 J