A spring of 5 times 10^3 text{ Nm}^{-1} sprin | vedclass

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Question

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

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The work done $W$ required to stretch 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 $k = 5 	imes 10^3 	ext{ Nm}^{-1}$.Initial extension $x_1 = 10 	ext{ cm} = 0.1 	ext{ m}$.Final extension $x_2 = 10 	ext{ cm} + 10 	ext{ cm} = 20 	ext{ cm} = 0.2 	ext{ m}$.Substituting the values:$W = \frac{1}{2} 	imes 5 	imes 10^3 	imes [(0.2)^2 - (0.1)^2]$$W = 2500 	imes [0.04 - 0.01]$$W = 2500 	imes 0.03 = 75 	ext{ J}$.

$A$ spring of $5 \times 10^3 \text{ Nm}^{-1}$ spring constant is stretched initially by $10 \text{ cm}$ from its unstretched position. The work required to stretch it further by another $10 \text{ cm}$ is: (in $\text{ J}$)

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