Question

A wire 2m in length suspended vertically stretched by 10mm when mass of 10 kg is attached to the lower end. The elastic potential gain by the wire

Answer 1

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

Δx = 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

Answer 2

See the attachement.........

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