Question

a 4kg block is attatched to a vertical sspring constant 800n/m. the spring stretches 5cm down. how much elastic potential energy is stored in the system

Answer 1

The Potential energy stored in the system is 1 J

Explanation:

Given-

Mass, m = 4 kg

Spring constant, k = 800 N/m

Distance, x = 5cm = 0.05m

Potential energy, U = ?

We know,

Change in potential energy is equal to the work done.

So,

$U = \frac{1}{2} k (x)^2$

By plugging in the values we get,

$U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J$

Therefore, Potential energy stored in the system is 1 J