How Do You Calculate Work Done on a Particle Given Its Position-Time Equation?

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To calculate the work done on a particle given its position-time equation, first determine the force acting on the particle by finding its acceleration, which is the second derivative of the position function. The position function x(t) = 4.1t - 0.64t² + 2.0t³ leads to the velocity and then acceleration equations. Work done (W) can also be expressed as the change in kinetic energy, which is the difference between the final and initial kinetic energy of the particle. The total work done from t = 0 to t = 8.1 s can be calculated by integrating the force over the displacement during that time interval. Understanding these relationships is crucial for solving the problem effectively.
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I was given the problem:

A single force acts on a 3.6 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.1t - 0.64t2 + 2.0t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.1 s.

I know that W= integral of Fx, but how do I find Fx?
 
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If ##x(t) = 4.1t - 0.64t^2 + 2.0t^3##, what else can you find about the motion, as well as the displacement?

I know that W= integral of Fx
That's one way to solve the problem, but what else is W equal to?
 

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