Under the action of a force a 2 kg body moves such that its position x in meters as a function of time t is given by x=(t4/4)+3 . Then work done by the force in first two seconds is

Question

Under the action of a force a 2 kg body moves such that its position x in meters as a function of time t is given by x=(t4/4)+3 . Then work done by the force in first two seconds is (A) 6 J (B) 10 J (C) 7 J (D) 64 J

Tardigrade Admin · Accepted Answer

x=(t4/4)+3 (d x/d t)=(4 t3/4)+0 v=t3 Using work energy theorem w=Δ k=(1/2)m(v (2))2-(1/2)m(v (0))2 =(1/2)(2)[(23)2 - 03] =64 J

Q. Under the action of a force a $2 k g$ body moves such that its position $x$ in meters as a function of time $t$ is given by $x = \frac{t ^{4}}{4} + 3$ . Then work done by the force in first two seconds is

$6 J$

$10 J$

$7 J$

$64 J$

$x = \frac{t ^{4}}{4} + 3$
$\frac{d x}{d t} = \frac{4 t ^{3}}{4} + 0$
$v = t^{3}$
Using work energy theorem
$w = Δ k = \frac{1}{2} m (v (2))^{2} - \frac{1}{2} m (v (0))^{2}$
$= \frac{1}{2} (2) [(2^{3})^{2} - 0^{3}]$
$= 64 J$

Q. Under the action of a force a 2kg body moves such that its position x in meters as a function of time t is given by x=4t4​+3 . Then work done by the force in first two seconds is

6J

10J

7J

64J