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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.

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

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Initial stretch of string 

x_{i}=5\ cm=0.05\ m

final stretch of string 

x_{f}=(5+5)\ cm=10\times 10^{-2}m = 0.1\ m

Work required = final potential energy - initial potential energy

W=\frac{1}{2}\times 5\times 10^{3}\times (0.1)^{2}-\frac{1}{2}\times 5\times 10^{3}\times (0.05)^{2}

W = 18.75\ J

Posted by

Vishwash Tetarwal

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