A 0.530 kg glider on an air track is attached to the end of an ideal spring with

Question

A 0.530 kg glider on an air track is attached to the end of an ideal spring with force constant 457 N/m ; it undergoes simple harmonic motion with an amplitude of 4.90×102 m .

Part A

Calculate the maximum speed of the glider.

Express your answer to three significant figures.

Part B

Calculate the speed of the glider when it is at x = 1.30×102 m .

Express your answer to three significant figures.

Part C

Calculate the magnitude of the maximum acceleration of the glider.

Express your answer to three significant figures.

Part D

Calculate the acceleration of the glider at x = 1.30×102 m .

Express your answer to three significant figures.

Part E

Calculate the total mechanical energy of the glider at any point in its motion.

Express your answer to three significant figures.

A 0.530 kg glider on an air track is attached to the end of an ideal spring with force constant 457 N/m ; it undergoes simple harmonic motion with an amplitude of 4.90×102 m .

Part A

Calculate the maximum speed of the glider.

Express your answer to three significant figures.

Part B

Calculate the speed of the glider when it is at x = 1.30×102 m .

Express your answer to three significant figures.

Part C

Calculate the magnitude of the maximum acceleration of the glider.

Express your answer to three significant figures.

Part D

Calculate the acceleration of the glider at x = 1.30×102 m .

Express your answer to three significant figures.

Part E

Calculate the total mechanical energy of the glider at any point in its motion.

Express your answer to three significant figures.

Explanation / Answer

A) Max kinetic energy=Max potential energy

0.5*m*V^2=0.5*k*A^2

V=sqrt(k/m) * A

V=1.439 m/s

B) Total energy remains the same

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

v=1.387 m/s

C) Ams acceleration = kA/m

=457*0.049/0.53

=42.251 m/s^2

d) Acceleration= 0.013*457/0.53

=11.209 m/s^2

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