when a mass of 2.5kg is hung on a steel wire, its length increases by 0.25cm . determine the work done to stretch the wire ?
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Explanation:
w=fs
f=weight=mg
2.5×10×0.0025
0.625
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Answer
Δx = stretch in the length of wire = 0.5 cm = 0.005 m
m = mass attached to the wire = 5 kg
g = acceleration due to gravity = 9.8 m/s²
W = weight attached to the wire
weight of mass hanged is given as
W = mg
W = 5 x 9.8
W = 49 N
k = force constant of the wire
the spring force by the wire balances the weight here
hence , Spring force = weight
k Δx = mg
k (0.005) = 49
k = 9800 N/m
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