A 1 kg cube is let go from rest and moves down a frictionless ramp that has a height of 0.2 m. At the ramp's bottom, the cube moves along a horizontal surface until it hits a spring that has a constant of k = 30 N/m. What is the spring's maximum compression?

The cube's GPE is 2 J.
Elastic PE = .5kx^2
2 = .5*30x^2
The max compression is .4 m.

assumeing no friction on the horizontal surface.

GPE=mg*.2=9.8*.2 J
KE=GPE=1/2 kx^2
x^2=2*9.8*.2/30= I dont get your answer.

1.96 = 1/2kx^2

3.92 = kx^2
3.92 = 30x^2
x^2 = 0.13
x = 0.36 m
Is that right?

E = m g h = 1*9.81*.2 = 1.96 yes

so
(1/2) k x^2 = 1.96
30 x^2 = 3.924
x^2 = .1308
x = 0.36 agreed

To find the maximum compression of the spring, we need to equate the potential energy of the cube at the top of the ramp to the potential energy stored in the compressed spring.

First, let's find the potential energy of the cube at the top of the ramp. The potential energy (PE) is given by the formula PE = m*g*h, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

PE = 1 kg * 9.8 m/s^2 * 0.2 m = 1.96 J

So, the potential energy of the cube at the top of the ramp is 1.96 J.

Next, let's find the potential energy stored in the compressed spring. The formula for elastic potential energy (PE) is given by the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the compression or expansion of the spring.

We need to solve the equation 0.5 * k * x^2 = PE.

Substituting the values, we have 0.5 * 30 N/m * x^2 = 1.96 J.

Simplifying the equation, we get:

15x^2 = 1.96

Dividing both sides by 15, we have:

x^2 = 1.96/15

Taking the square root of both sides, we find:

x = sqrt(1.96/15) = 0.231 m

Thus, the maximum compression of the spring is approximately 0.231 meters (or 0.23 m when rounded to two decimal places).