A 50 gram block is placed on a frictionless ramp 2 meters above the ground. It slides down the ramp and compresses a 140 N/m spring at the bottom. How far will the spring be compressed when the block is stopped by the spring?

say 1.7 m

Use conservation of energy. The loss of gravitational potential energy will equal the potential energy stored in the spring.

0.05 kg * 9.8 m/s^2 * 2 meters = (1/2) k X^2
X^2 = 0.014m^2
X = 0.12 m

I have neglected the gravitational potential energy change during the small spring compression . To include that effect, I would have to know the ramp angle

To determine how far the spring will be compressed when the block is stopped, we need to calculate the potential energy of the block as it slides down the ramp and then convert that energy into the spring's potential energy.

1. Find the potential energy of the block:
The potential energy of the block is given by the formula: potential energy = mass * gravity * height.

mass = 50 grams = 0.05 kg (converting grams to kg)
gravity = 9.8 m/s^2 (acceleration due to gravity)
height = 2 meters

potential energy = 0.05 kg * 9.8 m/s^2 * 2 meters
potential energy = 0.98 Joules

2. Convert the potential energy of the block into the potential energy of the spring:
The potential energy of the spring is given by the formula: potential energy = (1/2) * k * x^2.

k = 140 N/m (spring constant)
x = distance the spring is compressed (what we need to find)

0.98 Joules = (1/2) * 140 N/m * x^2

Now, rearrange the equation to solve for x:

x^2 = (2 * 0.98 Joules) / (140 N/m)
x^2 = 0.014 Joules / N
x^2 = 0.014 J/N

Taking the square root of both sides of the equation, we find:

x = √(0.014 J/N)

Calculating this gives us:

x ≈ 0.1187 meters or 11.87 cm

Therefore, the spring will be compressed approximately 0.1187 meters (or 11.87 cm) when the block is stopped by the spring.