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5.Work, Energy, Power and Collision
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The work done in joules in increasing the extension of a spring of stiffness $10\, N/cm$ from $4\, cm$ to $6\, cm$ is:
A
$1$
B
$10$
C
$50$
D
$100$
Solution
Given Spring stiffness $K=10 N / c m=1000 N / m,$ Initial Position $x_{1}=4 c m=0.04 m$ Final position $x_{2}=6 \mathrm{cm}=0.06 \mathrm{m}$
Work done by spring $=\frac{1}{2} K\left(x_{2}^{2}-x_{1}^{2}\right)$
$\Rightarrow$ Work done $=\frac{1}{2} \times 1000 \times\left(0.06^{2}-0.04^{2}\right)=1.0 \mathrm{J}$
Standard 11
Physics
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