A 0.550 kg glider on an air track is attached to the end of an ideal spring with

Question

A 0.550 kg glider on an air track is attached to the end of an ideal spring with force constant 456 N/m ; it undergoes simple harmonic motion with an amplitude of 4.40×102 m . Part A: Calculate the maximum speed of the glider. Part B: Calculate the speed of the glider when it is at x = 1.30×102 m . Part C: Calculate the magnitude of the maximum acceleration of the glider. Part D: Calculate the acceleration of the glider at x = 1.30×102 m . Part E: Calculate the total mechanical energy of the glider at any point in its motion.

Explanation / Answer

here,

mass , m = 0.55 kg

k = 456 N/m

amplitude , A = 0.044 m

A)

angular speed w = sqrt(k/m)

w = sqrt(456/0.55)

w = 28.79 rad/s

maximum speed , vm = A * w

v = 1.27 m/s

B)

x = - 0.013 m

using conservation of energy

0.5 * m * vm^2 = 0.5 * m * v^2 + 0.5 * k * x^2

0.55*1.27^2 = 456*0.013^2 + 0.55*v^2

v = 1.21 m/s

c)

the maximum accelration , am = A * w^2

am = 36.47 m/s^2

d)

the accelration , a = k * x /m

a = 456 * 0.013 /0.55

a = 10.78 m/s^2

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