# Physics/Math

posted by Technoboi11

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.

* Physics/Math - bobpursley, Thursday, March 1, 2007 at 6:28am

find the position at at t=2 and t=0. The difference in those is the displacement.

Work= force times displacement.

* Physics/Math - Technoboi11, Friday, March 2, 2007 at 12:11am

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

This is the answer that I got but it is an incorrect answer. Is there something wrong with my calculations???

x= 3.0t - 4.0t^ 2 + 1.0t^3
dx/dt= 3 - 8 t + 3t^2

d"x/dt"= acceleration= -8 + 6 t

F= ma = m (-8+6t)
INT F dx= work
work = INT m (-8+6t)(3 - 8 t + 3t^2)dt

integrate from t=0 to t= 2.

check my thinking.

Can you use W=1/2mv1^2-1/2mv2^2
Use 3-8t+3t^2 and solve for v1 when t=0 and v2 when t=2. Then plug v1 and v2 into the above equation with 5 as mass?

I used the above equation and it worked

1. PhysicsGuy

Use your last equation. Do not use d"x/dt" because this force is a variable force, and F = ma is not applicable here, I think.

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