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Best answer and 7 points guaranteed if u find answer.A spring of force constant 800 n/m has extension of 5 cm. find work done in extending from 5 cm to 15




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K = 800 N/m
x₁ = 5 cm = 0.05 m
x₂ = 15 cm = 0.15 m

Energy stored in a spring when it is elongated by a distance x =  \frac{1}{2}kx^2

Workdone = change\ in\ energy\ of\ spring\\ \\ W = \frac{1}{2}kx_2 ^{2}- \frac{1}{2}kx_1 ^{2}\\ \\W= \frac{1}{2}k(x_2^2-x_1^2) \\ \\W=\frac{1}{2}*800*(0.15^2-0.05^2) \\ \\W=400*(0.0225-0.0025)\\ \\W=400*0.02=\boxed{8\ J}
1 5 1
pls post it
In formula x square should b written first as we always do
Fina energy - initial energy
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