A force acts on a 3g 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 meters and t is in seconds. The work done during the first 4 second is :
2.88 J
450 mJ
490 mJ
530 mJ
A
450 mJ
B
2.88 J
C
530 mJ
D
490 mJ
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Solution
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We have,
mass, m=3g=0.003kg
x=3t−4t2+t3
Now,
v=dxdt=3−8t+3t2⇒dx=(3−8t+3t2)dt
⇒a=dvdt=0−8+6t
Now,
dw=Fdx
⇒dw=(ma)dx
⇒dw=(0.003)(−8+6t)(3−8t+3t2)dt
⇒dw=(0.003)(18t3−72t2+82t−24)dt
⇒w=(0.003)∫40(18t3−72t2+82t−24)dt
⇒w=0.003×176=0.528J
⇒w=530mJ
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