Question

A force acts on a 2 kg object so that its position is given as a function of time as $x = 3{t^2} + 5$. What is the work done by this force in the first 5 seconds?
A. 850 J
B. 950 J
C. 875 J
D. 900 J

Answer 1

We are given the mass of the object and its position. From position, we can calculate the acceleration of the mass and obtain force applied by using Newton’s second law. Then the work done by the force is equal to the product of force applied and the displacement of the object.
Formula used:
Newton’s second law is given as
$F = ma$
The work done by a force is given as
$W = F.d$

Complete answer:
We are given an object which mass is given as
$m = 2kg$
Its position is given to us as a function of time by the following expression.
$x = 3{t^2} + 5$
We need to find out the work done by a force applied on this mass in a duration of $t = 5s$. According to Newton’s second law of motion, the force applied on a body is equal to the product of mass and acceleration of the body. We have the value of mass and we just need to find out the acceleration of the mass. It can be calculated from the position of the object in the following way.
The acceleration is equal to the second order time derivative of position of the given object. So, we can write
$a = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{{d^2}}}{{d{t^2}}}\left( {3{t^2} + 5} \right) = \dfrac{d}{{dt}}\left( {6t} \right) = 6m/{s^2}$
Therefore, the force applied on the given object is given as
$F = ma = 2 \times 6 = 12N$
$x\left( 5 \right) = 3{\left( 5 \right)^2} + 5 = 3 \times 25 + 5 = 80m \\\ x\left( 0 \right) = 3{\left( 0 \right)^2} + 5 = 5m \\\$
The work done by this force is equal to the product of force applied and the displacement produced in the object. We have the value of force and the displacement is equal to the difference between the final position after $t = 5s$ and the initial position at $t = 0$. It can be calculated as follows:
The displacement is given as
$d = x\left( 5 \right) - x\left( 0 \right) = 80 - 5 = 75m$
Therefore, the work done can be calculated in the following way.
$W = F.d = 12 \times 75 = 900J$
This is the required value of work done.

Hence, the correct answer is option D.

Note:
There is another method by which we can arrive at the same result. That is by using the work energy theorem, where the work done by external force is equal to difference between final kinetic energy after 5 seconds and initial kinetic energy at t = 0. In that case, we just need to calculate velocity from the given expression for the position of the particle.