Question

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 $x=3t-4t^2+t^3$, where x is in metres 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

Answer 1

Answer: (A)

Step 1: Given that:

The mass(m) of the particle= $30g=\frac{30}{1000}kg=0.03kg$

The position of the particle is given by the function

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

Step 2: Calculation of the velocity of the particle:

The velocity of the particle in differential form is given by;

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

Thus,

$v=\frac{d}{dt}(3t-4t^2+t^3)$

$v=3-8t+3t^2$

The velocity of the particle at time $t=0$ s

$v_0=3-8\times 0+3{\left(0\right)}^2$

$v_0=3m{s}^{-1}$

The velocity of the particle after 4sec, will be

$v=3-8\times 4sec+3\times {\left(4s\right)}^2$

$v=3-32+48$

$v=19m{s}^{-1}$

Step 3: Calculation of work done by the particle:

According to the work-energy theorem,

Work done = Change in kinetic energy of the particle

Thus,

$W=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2$

$W=\frac{1}{2}m(v^2-v_0^2)$

Therefore, putting the values

$W=\frac{1}{2}\times 0.03kg\times \{{\left(19m{s}^{-1}\right)}^2-{\left(3m{s}^{-1}\right)}^2\}$

$W=\frac{1}{2}\times 0.03\times \{361-9\}$

$W=\frac{1}{2}\times 0.03\times 352$

$W=0.03\times 176$

$W=5.28J$

Thus,

The work done by the particle during the first 4sec is 5.28J.