Question

A wire 2 m in length suspended vertically stretches by. 10 mm when mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is (take g = 10 m//s^(2))

Answer 1

m = mass attached to lower end of the wire = 10 kg

Δ = stretch in the length of the wire due to weight of mass attached = 10 mm = 0.01 m

k = spring constant of the wire = ?

Using equilibrium of force in the wire

spring force by wire = weight of mass attached

k Δx = mg

inserting the above values

k (0.01) = (10 x 9.8)

k = 9800 N/m

Elastic potential energy is given as

U = (0.5) k Δx²

inserting the values

U = (0.5) (9800) (0.01)²

U = 0.49 J