A single force acts on a 1.8 kg particle-like object in such a way that the position of the object as a function of time is given by $x = 4.1t - 2.8t^2 + 3.3t^3$, with $x$ in meters and $t$ in seconds. Find the work done on the object by the force from $t = 0$ to $t = 6.0$ s.
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1t + 2.8t^2 + 3.3t^3. Acceleration (a) = d^2x/dt^2 = -5.6 + 19.8t Show more…
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