A cylindrical spring is found to elongate by 3 cm when stretched by a force of 500N.
Determine:
a) its force constant
b) the work done by a 600N force in elongating the spring
a) "F=k\\Delta x\\to k=F\/\\Delta x=500\/0.03=16667\\ (N\/m)"
b) "W=PE=\\frac{k(\\Delta x)^2}{2}=\\frac{k(F\/k)^2}{2}=\\frac{16667\\cdot(600\/16667)^2}{2}=10.8\\ (J)"
Comments
Leave a comment