work = integral f dx = change in kinetic energy if this is the only force
v =dx/dt = 4.1 -7.2 t + 13.5 t^2
at t = 0, v = 0
at t = 8.6
v = 4.1 - 7.2(8.6) + 13.5 (8.6)^2
= 941 m/s^2
Ke = .5 m v^2 = .5(1.1)(942)^2
= 486,642 Joules
A single force acts on a 1.1 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.1t - 3.6t2 + 4.5t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.6 s.
1 answer
1 answer