If a spring has a shiffness of 9,50nm-1 what work will be done in exered the spring by 60mm.
since F = 9.5 N/m, you want
W = ∫[0,0.06] 9.5 x dx = 0.0171Nm = 0.0171J
see
http://www.intmath.com/applications-integration/7-work-variable-force.php
W=1/2ke^2
K=950n/m
E=60mm=0.06m
W=950*(0.06)=57N
To determine the work done on a spring, you can use the formula:
Work = 0.5 * k * x^2
Where:
- Work is the amount of work done on the spring (in joules).
- k is the stiffness constant or spring constant (in newtons per meter, N/m).
- x is the displacement or elongation of the spring (in meters).
In this case, you are given that the stiffness of the spring (k) is 9.50 N/m, and the displacement (x) is 60 mm, which is equivalent to 0.06 meters.
Now, substitute the given values into the formula:
Work = 0.5 * 9.50 N/m * (0.06 m)^2
Simplifying the equation:
Work = 0.5 * 9.50 N/m * 0.0036 m^2
Work = 0.5 * 0.0342 N·m
Therefore, the work done on the spring is approximately 0.0171 joules.