(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

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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 $k = 5 \times 10^3 \, N/m$.Initial extension $x_1 = 5 \, cm = 0.05 \, m$.Final extension $x_2 = 5 \, cm + 5 \, cm = 10 \, cm = 0.10 \, m$.Substituting the values into the formula:$W = \frac{1}{2} \times (5 \times 10^3) \times ((0.10)^2 - (0.05)^2)$$W = \frac{1}{2} \times 5000 \times (0.01 - 0.0025)$$W = 2500 \times 0.0075$$W = 18.75 \, J$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Spring and Potential Energy of Spring

Medium

$A$ 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 .............. $J$.

AIEEE 2003 All AIEEE

A
$6.25$
B
$12.50$
C
$18.75$
D
$25.00$

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Work, Energy, Power and Collision

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Work Done by Spring and Potential Energy of Spring

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$A$ 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 .............. $J$.

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The potential energy of a long spring when stretched by $2\,cm$ is $U$. If the spring is stretched by $8\,cm$,the potential energy stored in it will be $.......\,U$.

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