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
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) To sketch the graph of the force against position, we need to plot the force (F) on the vertical axis and the displacement (d) on the horizontal axis. Since F = kd, where k is a constant, the graph will be a straight line passing through the origin (0,0) with a slope of k.
b) At position x, the value of the displacement is x. Substituting this value into the force equation F = kd, we get F = kx. Therefore, the value of the force at position x is kx.
c) The work done by a force is given by the formula: Work = Force × Displacement × cos(θ), where θ is the angle between the force and the displacement vectors. In this case, the angle between the force and displacement vectors is 0 degrees because they are parallel.
Using F = kx, we can rewrite the work expression as: Work = (kx) × (d) × cos(0) = kxd × 1 = kxd.
So, the expression for the work done by this force in terms of k and x is Work = kxd.