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.
For Further Reading
* 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=(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
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.