When the load applied to a wire is increased | vedclass

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Question

(A) The work done in stretching a wire is equal to the change in elastic potential energy stored in the wire.Elastic potential energy $U = \frac{1}{2}

Vedclass · Accepted Answer

$47 	imes 10^{-3} \ J$

Vedclass · Answer

(A) The work done in stretching a wire is equal to the change in elastic potential energy stored in the wire.Elastic potential energy $U = \frac{1}{2} F x$,where $F$ is the force and $x$ is the elongation.Given:Initial load $F_1 = 5 \ kg \ wt = 5 	imes 10 \ N = 50 \ N$,elongation $x_1 = 1 \ mm = 1 	imes 10^{-3} \ m$.Final load $F_2 = 8 \ kg \ wt = 8 	imes 10 \ N = 80 \ N$,elongation $x_2 = 1.8 \ mm = 1.8 	imes 10^{-3} \ m$.Work done $W = U_2 - U_1 = \frac{1}{2} F_2 x_2 - \frac{1}{2} F_1 x_1$.$W = \frac{1}{2} [(80 	imes 1.8 	imes 10^{-3}) - (50 	imes 1 	imes 10^{-3})]$.$W = \frac{1}{2} [144 	imes 10^{-3} - 50 	imes 10^{-3}] = \frac{1}{2} [94 	imes 10^{-3}] = 47 	imes 10^{-3} \ J$.

When the load applied to a wire is increased from $5 \ kg$ wt to $8 \ kg$ wt,the elongation of the wire increases from $1 \ mm$ to $1.8 \ mm$. The work done during the elongation of the wire is (Acceleration due to gravity $= 10 \ m/s^2$)

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