# A spring (80 N/m ) has an equilibrium length of 1.00 m. .?

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A spring (80 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 1.8 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released.

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Part A

If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

Part B

If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

Part C

Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring’s equilibrium position, what is the coefficient of friction μk?

• Part A

If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

The first step is to calculate the potential energy of the spring. Use the following equation.

PE = ½ * k * d^2 = ½ * 80 * 0.5^2 = 10 J

The force that causes the block to decelerate is the component of its weight that is parallel to the inclined plane.

Weight = 1.8 * 9.8 = 17.64 N

Force parallel = 17.64 * sin 41

This is approximately 11.6 N. Let’s determine the work done by this force.

Work = 17.64 * sin 41 * 0.5 = 8.82 * sin 41

To determine the kinetic energy of the block at this position, subtract the work from the potential energy of the spring.

KE = 10 – 8.82 * sin 41

This is approximately 4.21 J. The block will continue sliding up the inclined plane until its velocity is 0 m/s.

Work = 17.64 * sin 41 * d

17.64 * sin 41 * d = 10 – 8.82 * sin 41

d = (10 – 8.82 * sin 41) ÷ 17.64 * sin 41

This is approximately 0.36 meter. The total distance is 0.5 meter plus this distance.

d = 0.5 + 10 – 8.82 * sin 41) ÷ 17.64 * sin 41

This is approximately 0.86 meter.

Part B

If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

This time, the spring and the component of its weight that is parallel to the inclined plane will cause the block to decelerate.

For the spring, work = ½ * 80 * d^2 = 40 * d^2

40 * d^2 + 17.64 * sin 41 * d = 10

40 * d^2 + 17.64 * sin 41 * d – 10 = 0

d = (-17.64 * sin 41 ± √[(17.64 * sin 41)^2 – 4 * 40 * -10]) ÷ 80

d = (-17.64 * sin 41 + √[(17.64 * sin 41)^2 + 1600]) ÷ 80

This distance is approximately 0.38 meter.

Part C

Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring’s equilibrium position, what is the coefficient of friction μk?

KE = 10 – 8.82 * sin 41

This is approximately 4.21 J. This is the kinetic energy at the equilibrium position. For the block to stop moving at this position, this must be equal to the work that is done by the friction force.

Ff = μ * 17.64 * cos 41

Work = 0.5 * μ * 17.64 * cos 41 = μ * 8.82 * cos 41

μ * 8.82 * cos 41 = 10 – 8.82 * sin 41

μ = (10 – 8.82 * sin 41) ÷ 8.82 * cos 41

This is approximately 0.63.I hope this is helpful for you.

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