A 1.150 kg air-track der is attached to each end of the track by two co springs.

Question

A 1.150 kg air-track der is attached to each end of the track by two co springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, X- 0.190 m Find the effective spring constant of the system 2.63 N /m 151-9335 The glider is now released from rest at x 0.190 m. Find the maximum x-acceleration of the glider. 4.35x10 m/s 2 151-9984 Find the x-coordinate of the glider at time t 0.450T, where Tis the period of the oscillation 1.8698 Calculate the angular velocity and the period, then the time at which you must find the position Find the kinetic energy of the glider at x 0.00 m 4.75x 10-2 J 151-3675 Other Views My general preferences on what is marked as NEW

Explanation / Answer

Here ,

let the spring constant is k

a)

as F = Keq*x

0.50 = Keq * 0.19

Keq = 2.63 N/m

the effective spring constant is 2.63 N/m

b)

maximum acceleration = maximum force/m

maximum acceleration = 0.50/1.150

maximum acceleration = 0.435 m/s^2

the maximum acceleration of the block is 0.435 m/s^2

c)

as x = A*cos(2*pi*t/T)

for x = 0.19 * cos(2*pi*0.450)

x = -0.181 m

the x-position is -0.181 m

d)

at x = 0 m

kinetic energy of glider = max potential energy stored in springs

kinetic energy of glider = 0.5 * 2.63 * 0.19^2

kinetic energy of glider = 4.75 *10^-2 J

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