Gujarati
Hindi
5.Work, Energy, Power and Collision
medium

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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