A 395.0 g block is dropped onto a vertical spring with a spring constant k = 252.0 N/m. The block becomes attached to the spring, and the spring compresses 0.29 m before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?

What work is done by the spring?
What was the speed of the block just before it hit the spring?

See problem right above this one.

To answer these questions, we can apply the principles of work and energy. Let's break it down step by step:

1. Work done by the block's weight:
The work done by a force can be calculated using the equation:
Work = Force × Distance × cos(θ)
In this case, the force is the weight of the block, W = m × g (where m is the mass and g is the acceleration due to gravity). The distance is the vertical displacement of the block, which is equal to the compression of the spring, x = 0.29 m. The angle between the force and the displacement is 0 degrees since the force is acting vertically downward. Therefore, cos(θ) = 1.

Plugging in the values, we get:
Work = (m × g) × x × cos(θ) = (395.0 g) × (9.8 m/s^2) × (0.29 m) × (1) = 11.9 J

So the work done by the block's weight is 11.9 Joules.

2. Work done by the spring:
The work done by a spring can be calculated using the equation:
Work = (1/2) × k × x^2
Where k is the spring constant and x is the displacement of the spring compressed.

Plugging in the values, we get:
Work = (1/2) × (252.0 N/m) × (0.29 m)^2 = 10.468 Joules

So the work done by the spring is approximately 10.468 Joules.

3. Speed of the block just before hitting the spring:
To find the speed of the block just before it hits the spring, we need to use the principle of conservation of mechanical energy.
Since the block is dropped, initially its only form of energy is gravitational potential energy. At the bottom of its freefall, this potential energy is converted into kinetic energy just before it hits the spring.

Equating the gravitational potential energy of the block at the height it was dropped to the kinetic energy just before hitting the spring:
Potential energy = Kinetic energy
m × g × h = (1/2) × m × v^2
where h is the distance the block was dropped and v is the velocity of the block just before hitting the spring.

Simplifying the equation:
g × h = (1/2) × v^2
v^2 = 2 × g × h
v = √(2 × g × h)

Plugging in the values, we get:
v = √(2 × 9.8 m/s^2 × 0.29 m) = √(5.696) m/s ≈ 2.39 m/s

So the speed of the block just before hitting the spring is approximately 2.39 m/s.