SATHEE: Work Energy And Power Question 216

Work Energy And Power Question 216

Question: The position of a particle of mass 4 g, acted upon by a constant force is given by $x = 4 t^{2} + t$ , where x is in metre and t in second. The work done during the first 2 seconds is

Options:

A) $128 m J$

B) $512 m J$

C) $576 m J$

D) $144 m J$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] here, $m = 4, g = 4 \times 1 0^{- 3} k g$

$x = 4 t^{2} + t ∴ \frac{d x}{d t} = 8 t + 1 \frac{d^{2} x}{d t^{2}} = 8$

Work done. $W = \int f d x = \int m \frac{d^{2} x}{d t^{2}} (\frac{d t}{d t}) d t$

$= \int_{0}^{2} (4 \times 10^{- 3}) (8) (8 t + 1) d t$

$= 32 \times 10^{- 3} \int_{0}^{2} (8 t + 1) d t = 32 \times 1 0^{- 3} {[\frac{8 t^{2}}{2} + t]}^{2}_{0}$

$= 32 \times 1 0^{- 3} [4 {(2)}^{2} + 2 - 0] = 576 m J$

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Work Energy And Power Question 216

Question: The position of a particle of mass 4 g, acted upon by a constant force is given by x=4t2+t , where x is in metre and t in second. The work done during the first 2 seconds is

Options:

Answer:

Solution:

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Question: The position of a particle of mass 4 g, acted upon by a constant force is given by $x = 4 t^{2} + t$ , where x is in metre and t in second. The work done during the first 2 seconds is