You did not state in which direction the force acts. I assume in the direction of displacement.
Work done will be the area under the graph FvsD.
sketch the graph (F=kd is a straightline)
c) Using calculus, work=integral f.dx=
= int kx dx= 1/2 x^2
You should get this by measuring the area on your graph.
A variable force F acts through a displacement d. The magnitude of the force is proportional to the displacement, so F=kd, where k is constant.
a) Sketch a graph of this force against position up to position x.
b) According to the equation, what is the value of the force at x?
c) Determine an expression for the work done by this force in terms of k and x
1 answer
1 answer