Work done on the object problem

[norcal]

Homework Statement

A single force acts on a 5.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 2.0 s.

Homework Equations

x=3.0t-4.0t^2+1t^3
W=Fd
W=integral(Fdx)

The Attempt at a Solution

So when t=0, x=0.
When t=2, x=-2
displacement= -2-0 = -2m

W=Fd
F=ma=(5)(9.8)=49

W=(49 N)(-2 m)
W= -98 J

 

[Doc Al]

W=Fd
F=ma=(5)(9.8)=49

No. What makes you think that the acceleration equals 9.8? Is the acceleration even constant?

Hint: Consider the change in KE of the object.