a spring with spring constant of 30 N/m is stretched 0.2 m from its equilibrium position.
How much work must be done tostretch it an additional 0.1 m?
The Work required equals the potential energy increase:
W = (k/2)[(0.3)^2 - (0.2)^2],
where k is the spring constant.
W = 15*(0.05) = 0.75 joules
To calculate the work done to stretch the spring an additional 0.1 m, you can use the formula:
W = (1/2) * k * (x^2 - xo^2)
Where:
W is the work done
k is the spring constant
x is the final displacement
xo is the initial displacement
Given:
k = 30 N/m (spring constant)
x = 0.2 m (initial displacement)
xf = 0.2 m + 0.1 m = 0.3 m (final displacement)
Now, substitute the values into the formula:
W = (1/2) * 30 N/m * (0.3 m^2 - 0.2 m^2)
W = (1/2) * 30 N/m * (0.09 m^2 - 0.04 m^2)
W = (1/2) * 30 N/m * 0.05 m^2
W = 15 N/m * 0.05 m^2
W = 0.75 Nm
Therefore, to stretch the spring an additional 0.1 m, the work done is 0.75 Nm.
To find the work done to stretch the spring an additional 0.1 m, we can use the formula for work done on a spring:
Work = (1/2) * k * (x^2)
Where:
- Work is the amount of work done on the spring (in joules)
- k is the spring constant (in newtons per meter)
- x is the displacement from the equilibrium position (in meters)
Given in the question:
- Spring constant (k) = 30 N/m
- Initial displacement (x) = 0.2 m
Step 1: Calculate the work done for the initial displacement of 0.2 m
Work = (1/2) * k * (x^2)
Work = (1/2) * 30 N/m * (0.2 m)^2
Work = (1/2) * 30 N/m * 0.04 m^2
Work = 0.6 joules
Step 2: Calculate the work done for the additional displacement of 0.1 m
To find the additional work done, we need to calculate the new displacement by adding the initial displacement (0.2 m) with the additional displacement (0.1 m).
New displacement (x) = 0.2 m + 0.1 m
New displacement (x) = 0.3 m
Now, we can use the same formula to find the work done for the new displacement:
Work = (1/2) * k * (x^2)
Work = (1/2) * 30 N/m * (0.3 m)^2
Work = (1/2) * 30 N/m * 0.09 m^2
Work = 1.35 joules
Therefore, the total work done to stretch the spring an additional 0.1 m is 1.35 joules.