The work done in stretching a spring of force | vedclass

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Question

(C) The work done $W$ in stretching a spring from an initial extension $x_1$ to a final extension $x_2$ is given by the formula: $W = \frac{1}{2}k(x_2

Vedclass · Accepted Answer

$18.75$

Vedclass · Answer

(C) The work done $W$ in stretching a spring from an initial extension $x_1$ to a final extension $x_2$ is given by the formula: $W = \frac{1}{2}k(x_2^2 - x_1^2)$ Given: Force constant $k = 5 	imes 10^3 \ N/m$ Initial extension $x_1 = 5 \ cm = 5 	imes 10^{-2} \ m$ Final extension $x_2 = 10 \ cm = 10 	imes 10^{-2} \ m$ Substituting the values: $W = \frac{1}{2} 	imes (5 	imes 10^3) 	imes [(10 	imes 10^{-2})^2 - (5 	imes 10^{-2})^2]$ $W = \frac{1}{2} 	imes 5000 	imes [100 	imes 10^{-4} - 25 	imes 10^{-4}]$ $W = 2500 	imes 75 	imes 10^{-4}$ $W = 2500 	imes 0.0075 = 18.75 \ J$ (or $N-m$).

The work done in stretching a spring of force constant $5 \times 10^3 \ N/m$ from $5 \ cm$ to $10 \ cm$ is ...... $N-m$.

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