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