Question

A force given by F(x) = 4x^3 x̂ (in N/m^3) acts on a 1kg mass moving on a frictionless surface. The mass moves from x = 4 m to x = 7 m.
(a) How much work is done by the force?


(b) If the mass has a speed of 1 m/s at x = 4 m, what is its speed at x = 7 m?

Answer 1

work done = integral of F dx
= 4* integral from 4 to 7 of x^3 dx
= 4 * (7^4/4 - 4^4/4)
= 7^4 - 4^4
= 2145 Joules
=================
(1/2) m v^2 = 2145 + (1/2)m(1^2)

Answer 2

or in other words
final kinetic energy = work in + initial kinetic energy.