Question

A wire to metre in length suspended vertically stretches by 10 mm when mass of 10 kg is attached to the lower end the elastic potential energy gained by wire is

Answer 1

HEY!!

HERE IS YOUR ANSWER   :-  

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

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