A 0.560 kg glider on an air track is attached to the end of an ideal spring with

Question

A 0.560 kg glider on an air track is attached to the end of an ideal spring with force constant 458 N/m ; it undergoes simple harmonic motion with an amplitude of 4.60×102 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

Given Parameters

Part A

1.Force Constant ,K =458 N/m ;

2.Amplitude(Maximum displacment) = 4.60x10-2

4. Mass = 0.560 Kg

In simple Harmonic motion(SHM), we know that at maximum displacement kinetic energy is zero.

i.e , when x = A, v = 0,

so Energy is, E= 1/2 K A2

= 0.484 Joule

when, v = vmax, x=0, so

1/2 Mvmax2 = 0.484

vmax = 1.314 m/s

Part B

1/2 Mv2 +1/2 K x2 = E always

at x = 1.30×102 m

1/2x(0.560)v2 +1/2(458)(1.30×102)2 = 0.484

1/2x(0.560)v2 =0.445

v= 1.261 m/s

Part C

F = ma= -kx

maximum acceleration occurs when x(dislacement)is max that is at x = A

|max acceleration| = kA/m

   = 37.62 m/s2

Part D

ma = -Kx at any interval of time

a = -kx/m =10.63 m/s2

Part E

Total mechanical energy of the glider at any point in its motion

E = 0.484 Joule

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