SATHEE: Work Energy and Power - Result Question 10

Work Energy and Power - Result Question 10

12. A force acts on a $30 g m$ 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) $576 m J$

(b) $450 m J$

(d) $530 m J$

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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 $4 s e c$

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

Displacement in $4 s e c$

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

$∴$ Work $=$ Force $\times$ displacement

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

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

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Work Energy and Power - Result Question 10

12. A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3, where x is in metres andt isin seconds. The work doneduring the first 4 seconds is

Answer:

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12. A force acts on a $30 g m$ 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