work done problem (urgent)

1. since x = 3t - 4t^2 + t^3
hence, velocity v = dx/dt = 3 - 8t +3t^2
At t = 0 s, v(0) = 3 m/s
At t = 4 s, v(4) = [3 - 8x4 + 3x(4^2)] m/s = 19 m/s

Increase of kinetic energy = [(1/2).3.19^2 - (1/2).(3).(3^2)] J = 528 J
Because work-done = change of kinetic energy
hence, work-done = 528 J

39. As you have already calculated the the work-dones in (a) to (c), you should find the values respectively: 42 J, 30J and 12 J

Since the particle starts from rest, using work-done = change of kinetic energy
(d) At x = 4m,
42 = (1/2).2.v^2. where v is the speed at x = 4 m
i.e. v = 6.48 m/s

(e) At x = 7m,
30 = (1/2).2.v'^2 where v' is the speed at x = 7m
i.e. v' = 5.48 m/s

(f) At x = 9m,
12 = (1/2).2v"^2 where v" is the speed at x = 9 m
i.e. v" = 3.46 m/s

Because the net work-dones are all positive, the particle is moving in the +ve x direction at the three distances.

41. Work done = integrate[F.dx] = integrate[(cx - 3x^2).dx]
the limits of integration is from x = 0 m to x = 3 m

After integration,
work done = 9c/2 - 27
SInce it is given that the change in kinetic energy = (11 - 20) J = -9 J
thus, 9c/2 - 27 = -9
solve for c gives c = 4 N/m

initial kinetic energy=m/2*(dx/dt)=0
final kinetic energy=m/2*(dx/dt)^2=1093.5
work done = 1093.5J