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
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
To calculate the work done in stretching the spring, we can use the formula:
Work = (1/2) * k * x^2
Where:
k is the spring constant (27 N/m)
x is the displacement from the equilibrium position (0.065 m)
Let's substitute the values into the formula and calculate the work:
Work = (1/2) * 27 N/m * (0.065 m)^2
= 0.5 * 27 N/m * (0.065 m)^2
= 0.5 * 27 N/m * 0.004225 m^2
= 0.08536275 N * m
Therefore, the work done to stretch the spring an additional 0.065 m is approximately 0.085 J.
To determine the work required to stretch the spring an additional 0.065 m, we need to use the formula for the potential energy stored in a spring:
U = (1/2) k x^2
Where:
U is the potential energy stored in the spring,
k is the spring constant, and
x is the displacement from the equilibrium position.
First, let's calculate the initial potential energy U1 when the spring is stretched by 0.2 m:
U1 = (1/2) * (27 N/m) * (0.2 m)^2
U1 = (1/2) * 27 * 0.04
U1 = 0.54 J
Next, we calculate the final potential energy U2 when the spring is stretched by an additional 0.065 m:
U2 = (1/2) * (27 N/m) * (0.265 m)^2
U2 = (1/2) * 27 * 0.070225
U2 = 0.9488625 J
Finally, we can find the work done to stretch the spring an additional 0.065 m by taking the difference between the final and initial potential energies:
Work = U2 - U1
Work = 0.9488625 J - 0.54 J
Work = 0.4088625 J
Therefore, the work required to stretch the spring an additional 0.065 m is approximately 0.409 J.
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