A block of mass 0.688 kilograms resting on a smooth horizontal surface is attached to a spring with a spring constant of 33.4 newtons per meter. The block is pulled horizontally 14.1 centimeters from its equilibrium position and released. a. What is the total mechanical energy of the system? Include units in your answer. b. What is the speed of the block when it is 9.0 centimeters from its equilibrium position? Include units in your answer. c. One end of the spring is then attached to a support so it hangs vertically, and the block is suspended from it and lowered into a resting position. How far does the block stretch the spring? Include units in your answer. d. As shown below, the block is then lifted 14.1 centimeters (labeled d in the figure) and released. What is the speed of the block when it is 9.0 centimeters from its equilibrium position (labeled y in the figure)? Include units in your answer. More information. Hint: Since the vertical position of the block is changing, remember to include gravitational potential energy in the equation for the law of conservation of mechanical energy. REFER TO THE IMAGE FOR THIS QUESTION.

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A block of mass 0.688 kilograms resting on a smooth horizontal surface is attached to a spring with a spring constant of 33.4 newtons per meter. The block is pulled horizontally 14.1 centimeters from its equilibrium position and released.

a. What is the total mechanical energy of the system? Include units in your answer.

b. What is the speed of the block when it is 9.0 centimeters from its equilibrium position? Include units in your answer.

c. One end of the spring is then attached to a support so it hangs vertically, and the block is suspended from it and lowered into a resting position. How far does the block stretch the spring? Include units in your answer.

d. As shown below, the block is then lifted 14.1 centimeters (labeled d in the figure) and released. What is the speed of the block when it is 9.0 centimeters from its equilibrium position (labeled y in the figure)? Include units in your answer. More information. Hint: Since the vertical position of the block is changing, remember to include gravitational potential energy in the equation for the law of conservation of mechanical energy. REFER TO THE IMAGE FOR THIS QUESTION.

 

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Step 1

The block has a mass of 0.688 kg  and is initially at rest on a horizontal surface, and is attached to a spring with spring constant of 33.4 N/m

This block is then pulled horizontally to a distance of 14.1 cm and then released.

As soon as the block is released, it will start performing a simple harmonic motion(SHM), about its mean position, with a constant amplitude of 14.1 cm or 0.141 m

While performing oscillations, there is a mechanical energy stored in the block-spring system, given by the equation

E=12mω2A2ω is the angular speed of SHMA is the amplitude of SHM

The angular speed of oscillations of SHM is given as

ω=kmk is the spring constantm is the mass attached to the spring

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