A spring of spring constant 5 times 10^3 , N/ | vedclass

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(C) The work done 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

Vedclass · Accepted Answer

$18.75$

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(C) The work done 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:Spring constant $k = 5 	imes 10^3 \, N/m$.Initial extension $x_1 = 5 \, cm = 5 	imes 10^{-2} \, m$.Final extension $x_2 = 5 \, cm + 5 \, cm = 10 \, cm = 10 	imes 10^{-2} \, m$.Substituting the values:$W = \frac{1}{2} 	imes (5 	imes 10^3) 	imes [(10 	imes 10^{-2})^2 - (5 	imes 10^{-2})^2]$$W = 2500 	imes [100 	imes 10^{-4} - 25 	imes 10^{-4}]$$W = 2500 	imes [75 	imes 10^{-4}]$$W = 2500 	imes 0.0075 = 18.75 \, N-m$.

$A$ spring of spring constant $5 \times 10^3 \, N/m$ is stretched initially by $5 \, cm$ from the unstretched position. Then the work required to stretch it further by another $5 \, cm$ is .............. $N-m$.

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