If a spring had a stiffness of 950 Newton meter, what work will be done in extending the spring by 60milimeters square
Thanks a lot
To calculate the work done in extending the spring, we can use the formula for potential energy stored in a spring:
Elastic Potential Energy (PE) = (1/2) * k * x^2
Where:
- PE is the potential energy stored in the spring
- k is the stiffness or spring constant of the spring
- x is the displacement or extension of the spring from its equilibrium position
In this case, the stiffness (k) is given as 950 Newton meter and the extension (x) is 60 millimeters (which is 0.06 meters).
Substituting the given values into the formula, we have:
PE = (1/2) * 950 N/m * (0.06 m)^2
PE = (1/2) * 950 N/m * 0.0036 m^2
PE = 1.71 J (joules)
Therefore, the work done in extending the spring by 60 millimeters is 1.71 joules.
Good grief, read what you type!
Newtons PER meter
I assume you mean 60 millimeters, not square
In other words maybe:
k = 950 N/m
x = 0.060 meters = 6* 10^-2 m
if so then
W = (1/2) k x^2
= 475(36*10^-4) Joules
= 1.71 J
or perhaps you mean x^2 = 60 *10^-6 m^2 ?