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Q.

Position of a particle of mass 30g varies as $x = 3 t - 4 t^{2} + t^{3}$ where ‘x’ is in metre and ‘t’ is in seconds. Work done by the net force acting on the particle during the first 4 seconds is

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a

5.28 J

b

7.5 J

c

8.45 J

d

9.32 J

answer is A.

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Detailed Solution

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

$V = 3 - 8 t + 3 t^{2}$

V at t = 0 = 3 m/s
V at t = 4 sec = 19 m/s
Work done = $Δ K = \frac{1}{2} (\frac{30}{1000}) (19^{2} - 3^{2})$

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