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 :
A
2.88 J
B
450 mJ
C
490 mJ
D
530 mJ
Medium
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Updated on : 2022-09-05
Solution
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Correct option is D)
We have,
mass, m=3g=0.003kg
x=3t−4t2+t3
Now,
v=dtdx=3−8t+3t2⇒dx=(3−8t+3t2)dt
⇒a=dtdv=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)∫04(18t3−72t2+82t−24)dt
⇒w=0.003×176=0.528J
⇒w=530mJ
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