Q4.7 Variable force (S)
A force acts on a 3 kg object in such a way that the position of the object as a function of time is given byx(t)=3t-4t2+t3
Find the work done on the object by the force from t=0 to t=4 sSolution
Reveal
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![\[ v(t) = \frac{d x}{dt} = 3 - 8 t + 3 t^2 \]](mastermathpng-0.png)

Solution
The work done must be equal to the difference in kinetic energy at the initial and final times (Key Point 3.7). The velocity is given by:
![\[ v(t) = \frac{d x}{dt} = 3 - 8 t + 3 t^2 \]](mastermathpng-0.png)
Therefore v(0)=3 ms-1, and v(4)=19 ms-1. The work done on the object is therefore
![\[ W = \frac{1}{2} m \left(v_f^2 - v_i^2 \right) = 0.5 \times 3 \times (19^2 - 3^2) = 528\,J \]](mastermathpng-1.png)