When a 4 kg mass is hung vertically on a light spring that obeys hooke's law, the spring streches by 2cm. The work required to be done by an external agent in streching this spring by 5 cm will be?
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Hello buddy,
● Answer-
2.5 J
● Explanation-
# Given-
m = 4 kg
x = 2 cm = 2×10^-2 m
# Solution-
Here, restoring force balances weight of mass.
kx = mg
k = mg/x
k = 4 × 10 / 2×10^-2
k = 2000 N/m
Work done for x = 5 cm,
W = 1/2 kx^2
W = 1/2 × 2000 × (5×10^-2)^2
W = 2.5 J
Work done for 5 cm extension is 2.5 J.
Hope this helps...
● Answer-
2.5 J
● Explanation-
# Given-
m = 4 kg
x = 2 cm = 2×10^-2 m
# Solution-
Here, restoring force balances weight of mass.
kx = mg
k = mg/x
k = 4 × 10 / 2×10^-2
k = 2000 N/m
Work done for x = 5 cm,
W = 1/2 kx^2
W = 1/2 × 2000 × (5×10^-2)^2
W = 2.5 J
Work done for 5 cm extension is 2.5 J.
Hope this helps...
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