The natural length of spring is 0.3 m and its spring constant is 30 n/m. How much work is done by the applied external force to stretch the spring from 0.1 to 0.2 m?
Answers
SOLUTION.
HEY if we think deep about the work done by external agent than we find if F be the external force acts on the body.
Than the work done in moving the body through X is F.XCosθ
Where θ is the angle between the Force and displacement.
ACC to your question F is the spring force. And the angle between force and displacement is 180° . And important point is that force is variable because X varies. So we use differential form of the EQ
dw = F.dxcosθ
dw = k.x.dxcos(180°)
dw = -k.x.dx
∫dw = -k ∫x.dx
[W] = -k[x²/2]
Limits of work goes from 0 to w
Limits of displacement goes from 0.1 to 0.2
W - 0 = -k [ (0.2)²- (0.1)²]
W = -30 (0.04 - 0.01)
W = -30×0.03
W = -0.9joules
#answerwithquality
#BAL
answer is 0.9 joules
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