Asked by Alex
A spring is stretch 40mm by a force of 15N ...what's the work done
Answers
Answered by
Damon
x = 40 * 10^-3 meters
F = k x
15 = k (40 *10^-3)
k = 15/40 * 10^3 Newtons/meter = 3/8 * 10^3
PE = work done = (1/2) k x^2
= (1/2)(3/8*10^3)(1600*10^-6) Joules
F = k x
15 = k (40 *10^-3)
k = 15/40 * 10^3 Newtons/meter = 3/8 * 10^3
PE = work done = (1/2) k x^2
= (1/2)(3/8*10^3)(1600*10^-6) Joules
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