The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x=3t^2+4t+5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

The correct option is: D(75 J)

Step 1: Given

Distance$\left(x\right)=3t^2+4t+5$

Mass(m)= 0.5 kg

Time(t)= 2 s

F, v and a are force, velocity and acceleration respectively.

Step 2: Formula used

$v=\frac{dx}{dt}$

F=m*a

$a=\frac{dv}{dt}$

Work done(W)=Force(F)*Displacement(x)

Step 3: Finding force

Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}(3t^2+4t+5)=6t+4$

Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}(6t+4)=6m{s}^{-2}$
Therefore, applied force, $F=ma=0.5\times 6=3N$

Step 4: Finding displacement in 2 sec
Now, the distance moved in 2 s, $x=3\times (2{)}^2+(4\times 2)+5=25m$

Step 5: Finding work done

$\therefore$ Work done,$W=Fx=3\times 25=75J$
Hence, the correct choice is (D).