A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

Q. A force acts on a

30 g

particle in such a way that the position of the particle as a function of time is given by

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

, where

x

is in meters and

t

is in seconds. The work done during the first

4

seconds is

Q. A force acts on a

30

g particle in such a way that the position of the particle as a function of time is given by

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

, where x is in metre and t is in second. The work done during the first

4

second is?

Join BYJU'S Learning Program

A force acts on a 3 g 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 second. The work done during the first 4 second is:

The correct option is C 528 mJ Given, x=3t−4t2+t3 dx=d(3t−4t2+t3)=(3−8t+3t2)dt Accelerationa(x)=d2xdt=d2xdt=d(3−8t+3t2)dt=−8+6t Work done =dw=ma.dx=m(6t−8)(3−8t+3t2)dt ∫W0dw=m∫40(6t−8)(3−8t+3t2)dt W=m×176=0.003×176=528×10−3J W=528×10−3J

A force acts on a $3 g$ particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where $x$ is in meters and $t$ is in second. The work done during the first $4$ second is: