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Question

Physics Question on work

When a spring is stretched by a distance xx, it exerts a force, given by F=(−5x−16x3) NF = (-5x - 16x^{3})\, N. The work done, when the spring is stretched from 0.1 m0.1 \,m to 0.2 m0.2\, m is

A

8.7×10−2 J8.7 \times 10^{-2}\,J

B

12.2×10−2 J12.2 \times 10^{-2}\, J

C

8.7×10−1 J8.7 \times 10^{-1}\,J

D

12.2×10−1 J12.2 \times 10^{-1}\,J

Answer

8.7×10−2 J8.7 \times 10^{-2}\,J

Explanation

Solution

F=−5x−16x3=−(5+16x2)x=−kxF=-5x-16x^{3}=-(5+16x^{2})x=-kx ∴k=5+16x2\therefore k=5+16x^{2} Work done, W=12k2x22−12k1x12W=\frac{1}{2}k_{2}x^{2}_{2}-\frac{1}{2}k_{1}x^{2}_{1} =12[5+16(0.2)2](0.2)2=12[5+16(0.1)2](0.1)2=\frac{1}{2}[5+16(0.2)^{2}](0.2)^{2}=\frac{1}{2}[5+16(0.1)^{2}] (0.1)^{2} =2.82×4×10−2−2.58×10−2=2.82 \times 4 \times 10^{-2}-2.58\times 10^{-2} =8.7×10−2 J=8.7 \times 10^{-2}\,J