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=3t-4t2+t3 where x is in meters and t is in seconds. The work done during the first 4 seconds is

Question

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=3t-4t2+t3 where x is in meters and t is in seconds. The work done during the first 4 seconds is (A) 5.28 J (B) 450 mJ (C) 490 mJ (D) 530 mJ

Tardigrade Admin · Accepted Answer

v=(d x/d t)=3-8t+3t2 Speed at t=0 is v0=3 m / s Speed at t=4 is v4=19 m / s According to work energy theorem W=(1/2)m(v42 - v02) W=(1/2)× 0.03× (1 92 - 32)=5.28 J

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

$5.28 J$

$450 m J$

$490 m J$

$530 m J$

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$
Speed at $t = 0$ is
$v_{0} = 3 m / s$
Speed at $t = 4$ is
$v_{4} = 19 m / s$
According to work energy theorem
$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$
$W = \frac{1}{2} \times 0.03 \times (1 9^{2} - 3^{2}) = 5.28 J$

Q. A force acts on a 30g 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 seconds is

5.28J

450mJ

490mJ

530mJ