Work Energy and Power - Result Question 10

12. 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 andt isin seconds. The work doneduring the first 4 seconds is

[1998]

(a) $576 mJ$

(b) $450 mJ$

(c) $490 mJ$

(d) $530 mJ$

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Answer:

Correct Answer: 12. (a)

Solution:

  1. (a) $x=3 t-4 t^{2}+t^{3}$

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

Acceleration $=\frac{d^{2} x}{d t^{2}}=-8+6 t$

Acceleration after $4 sec$

$=-8+6 \times 4=16 ms^{-2}$

Displacement in $4 sec$

$=3 \times 4-4 \times 4^{2}+4^{3}=12 m$

$\therefore$ Work $=$ Force $\times$ displacement

$=$ Mass $\times$ acc. $\times$ disp.

$=3 \times 10^{-3} \times 16 \times 12=576 mJ$