the length of a steel wire increases by 0.5 cm when it is loaded with a weight of 5 kg. calculate
i) force constant
ii) work done in streching the wire
Answers
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
ii)
work done stretching the wire is given as
W = (0.5) k Δx²
W = (0.5) (9800) (0.005)²
W = 0.1225 J
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
ii)
work done stretching the wire is given as
W = (0.5) k Δx²
W = (0.5) (9800) (0.005)²
W = 0.1225 J