A spring of length 10.0 cm and spring constant 160 kN/m is compressed to a length of 4.00 cm. If, when released, it pushes a block of mass 100 kg, a) what is the speed of the block after it slides up a 15.0 cm hill? b) What is its speed after it falls into a 10.0 cm dip?

To solve this problem, we need to consider the conservation of mechanical energy. The initial potential energy stored in the compressed spring gets converted into the kinetic energy of the block as it slides up the hill.

Let's start by finding the potential energy stored in the compressed spring:
Potential energy (PE) = (1/2)kΔx^2
where k is the spring constant and Δx is the compression or extension of the spring.

Given:
Length of the spring, L = 10.0 cm = 0.10 m
Spring constant, k = 160 kN/m = 160,000 N/m
Compression, Δx = 4.00 cm = 0.04 m

Substituting the values:
PE = (1/2) * 160,000 * (0.04)^2
PE = 128 J

Now, let's calculate the speed of the block after it slides up the 15.0 cm hill.

Since the block slides up the hill, the potential energy will be converted into both kinetic energy and gravitational potential energy.

At the maximum height, all potential energy will be converted into kinetic energy, so:
Potential energy (PE) = Kinetic energy (KE)
mgh = (1/2)mv^2, where m is the mass of the block, g is the acceleration due to gravity, h is the height, and v is the velocity of the block.

Given:
Mass of the block, m = 100 kg
Height of the hill, h = 15.0 cm = 0.15 m
Acceleration due to gravity, g = 9.8 m/s^2

Substituting the values:
(mgh) = (1/2)mv^2
100 * 9.8 * 0.15 = (1/2) * 100 * v^2
14.7 = 50v^2
v^2 = 0.294
v = √0.294
v ≈ 0.542 m/s

Therefore, the speed of the block after it slides up the 15.0 cm hill is approximately 0.542 m/s.

For part (b), let's calculate the speed of the block after it falls into a 10.0 cm dip.

In this case, the potential energy will be converted into kinetic energy only.

Potential energy (PE) = (1/2)mv^2

Given:
Height of the dip, h = 10.0 cm = 0.10 m

Substituting the values:
(mgh) = (1/2)mv^2
100 * 9.8 * 0.10 = (1/2) * 100 * v^2
9.8 = 50v^2
v^2 = 0.196
v = √0.196
v ≈ 0.443 m/s

Therefore, the speed of the block after it falls into a 10.0 cm dip is approximately 0.443 m/s.