A spring with spring constant k is extended f | vedclass

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Question

(B) The restoring force exerted by the spring is given by $F = -kx$,where $k$ is the spring constant and $x$ is the displacement.To calculate the work

Vedclass · Accepted Answer

$\frac{1}{2}kx_1^2$

Vedclass · Answer

(B) The restoring force exerted by the spring is given by $F = -kx$,where $k$ is the spring constant and $x$ is the displacement.To calculate the work done by an external agent to extend the spring,we apply an external force $F_{ext} = -F = kx$.The work done $W$ is the integral of the force over the displacement:$W = \int_{0}^{x_1} F_{ext} dx = \int_{0}^{x_1} kx dx$$W = k \left[ \frac{x^2}{2} \right]_{0}^{x_1}$$W = k \left( \frac{x_1^2}{2} - 0 \right)$$W = \frac{1}{2} k x_1^2$Thus,the work done in extending the spring is $\frac{1}{2} k x_1^2$.

$A$ spring with spring constant $k$ is extended from $x = 0$ to $x = x_1$. The work done will be

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