- #1

Lebombo

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## Homework Statement

Suppose is a 10kg block sitting at rest on the ledge of a building 20 meters tall. The 10kg block applies a Force of F = ma = 10kg(10m/s^2) = 100N. The potential energy with reference to the ground would then be PE = mgh = 10kg(10m/s^2)(20m) = 2000J.

If the 10kg block is then gently placed on a 20m tall spring scale of spring constant k = 10N/M

How far does the spring compress?

Determine the amount of energy in the fully compressed spring.

How much force is the block applying to the spring when the spring is fully compressed?

## Homework Equations

F*d = mgh

F*d = (1/2)kx^2

F = kx

## The Attempt at a Solution

Answer to question 1)

F = kx

F/k = x

mg/k = x

(10kg)(10m/s^2) / (10N/m) = x

10 meters = x

So the spring compresses 10m when the 10kg block is gently placed on the spring.Answer to question 2)

Gravitational PE = F*d = mgh = (10kg)(10m/s^2) * 10m = 100N*10m/s^2 = 1000J

Spring PE = (1/2)kx^2 = (1/2)(10N/m)(10m)^2 = 500J

Answer to Question 3

1st guess:

kx = F = 100N2nd guess:

F*d = (1/2)kx^2

F(10m) = 500J

F = 500J/10m = 50N3rd guess:

(1/2)(base)(height) = (1/2)(Δx)(Δy) = (1/2)(Distance)(Force) = (1/2)(10m)(F) = 1000J

F = 2(1000J)/(10m) = 200NDon't know if any of this is correct. I don't fully understand the energy and force and whether or not potential energy even has any part in determining the energy and force. Or how spring potential energy determines the energy and force. Thank you for any reply.