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Q.

Force acts on a 20g particle in such a way that the position of the particle as a function of time is given by x=2t−3t2+2t3  where x is in metre and t in second. The work done in first three seconds is

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a

14.4 J

b

5.28 J

c

7.2 J

d

28.8 J

answer is C.

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Detailed Solution

x=2t−3t2+2t3

V=2−6t+6t2

Initial velocity V1=2m/s

velocity at t = 3 sec,V2=38m/s

Using work energy theorem,

W=Δk=12201000382-22=14.4J

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