Potential energy stored = work in
W = (1/2) k x^2
dW/dx = k x = Force
dW = k x dx
approximately delta W = k x delta x
delta W = 27 * .2 * .065 Joules
A spring with spring constant of 27 N/m is stretched 0.2 m from its equilibrium position. How much work must be done to stretch it
an additional 0.065 m? Answer in units of J
2 answers
Work done = Potential Energy change.
= (1/2)*k[0.2065^2 - 0.200)^2]
where k = 27 N/m
The answer will be in Joules if the stretch distance X is in meters, as above
= (1/2)*k[0.2065^2 - 0.200)^2]
where k = 27 N/m
The answer will be in Joules if the stretch distance X is in meters, as above
0 answers