1. A block of mass 0.257 kg is placed on top of a light, vertical spring of forc

Question

1. A block of mass 0.257 kg is placed on top of a light, vertical spring of force constant 4565 N/m and pushed downward so that the spring is compressed by 0.092 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)

2. A 1.44 kg object is held 1.27 m above a relaxed, massless vertical spring with a force constant of 333 N/m. The object is dropped onto the spring.

a) How far does the object compress the spring?

b) Repeat part (a), but now assume that a constant air-resistance force of 0.720 N acts on the object during its motion.

c)How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2 and air resistance is neglected?

3. A 237-g block is pressed against a spring of force constant 1.21 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions.

a)if the ramp exerts no friction force on the block

b) if the coefficient of kinetic friction is 0.414

Explanation / Answer

1)

If the statement means that the block compresses the spring a total of 0.092 m from a previously relaxed position,

mg = kx

(0.257 kg)(9.81 m/s²) = (4565 N/m)(x)

x = 5.423e-4 m

OK, this is pretty insignificant compared to the 0.092 m additional compression... to arrive at an answer needing 2 decimal places, so there's little need to account for this. So...

If this means that the block is placed on the spring, the spring compresses slightly due to the weight of the block, then the block is pushed downward to compress the spring another 0.092 m, this would be the calculation:

(initial potential energy of compressed spring) = (final gravitational potential energy of block)

(1/2)kd² = mgy

(1/2)(4565 N/m)(0.092 m)² = (0.257 kg)(9.81 m/s²)(y)

y = 7.66 m

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