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=3tβ4t2+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
Open in App
Solution
Verified by Toppr
Correct option is D)
We have,
mass, m=3Β g=0.003Β kg
x=3tβ4t2+t3
Now,Β
v=dtdxβ=3β8t+3t2βdx=(3β8t+3t2)dt
βa=dtdvβ=0β8+6t
Now,
dw=Fdx
βdw=(ma)dx
βdw=(0.003)(β8+6t)(3β8t+3t2)dt
βdw=(0.003)(18t3β72t2+82tβ24)dt
βw=(0.003)β«04β(18t3β72t2+82tβ24)dt
βw=0.003Γ176=0.528Β J
βw=530Β mJ
Solve any question of Work, Energy and Power with:-