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.
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* 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???
Working on this myself. Use the work energy theorem.
Your calculations seem to be correct, but there is a mistake in the given position equation. The equation x = 3.0t - 4.0t^2 + 1.0t^3 is incorrect. It should be x = 3.0t - 4.0t^2 + 1.0t^3/3. This is because the equation for position as a function of time for an object under constant force is given by x = at^2/2 + bt^3/3 + ct^4/4 + ... where a, b, c, ... are constants.
To find the correct work done on the object from t = 0 to t = 2.0 s, we need to find the displacement of the object during that time interval.
- First, we find the position at t = 2 s using the corrected equation:
x = 3.0t - 4.0t^2 + 1.0t^3/3
x = 3.0(2) - 4.0(2)^2 + 1.0(2)^3/3
x = 6 - 16 + 16/3
x = -4 + 16/3
x = -4 + 5.33
x = 1.33 m
- Next, we find the position at t = 0 s:
x = 3.0t - 4.0t^2 + 1.0t^3/3
x = 3.0(0) - 4.0(0)^2 + 1.0(0)^3/3
x = 0 - 0 + 0/3
x = 0
- The displacement during the time interval is the difference between the positions:
displacement = x(t=2) - x(t=0) = 1.33 - 0 = 1.33 m
- The work done on the object is given by the product of the force and displacement:
work = force * displacement
work = (5 kg) * (9.8 m/s^2) * (1.33 m)
work = 65.17 J
Therefore, the correct work done on the object by the force from t = 0 to t = 2.0 s is approximately 65.17 Joules.