A force acts on a 30 gm particle in such a way that the position of the particle

Question

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where x is in metres and t is in seconds. The work done during the first 4 seconds is

Options:

A . 5.28 J

B . 450 mJ

C . 490 mJ

D . 530 mJ

Answer: Option A
:
A

v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}

∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s

W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})

(According to work energy theorem)

= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J

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