Question

A body of mass $0.5$ kg travels in a straight line with velocity $v = a x^{\dfrac{3}{2}}$ where, $a = 5 \left(m^{\dfrac{1}{2}} s^{-1} \right)$. The work done by the net force during its displacement from $x = 0$ to $x = 2m$ is

Answer 1

Hint: Work shifts energy from one position to another. Work can convert the potential energy of an automatic device, the heat energy, or the electrical energy in an electrical device. The work and kinetic energy principle defines that the work produced by the sum of all forces operating on a particle equals the difference in the particle's kinetic energy. This explanation can be reached to rigid bodies by describing the work of the torque and rotational dynamic energy.

Complete step-by-step solution:
Given: velocity, $v = a x^{\dfrac{3}{2}}$
$a = 5$
Mass of body, $ m = 0.5 Kg$
Initial velocity of body at $x = 0$,
$v_{i} = 5 \times 0 = 0$
Final velocity of body at $x = 2$,
$v_{f} = 5 \times 2^{\dfrac{3}{2}} = 10\sqrt{2} ms^{-1}$
We will apply the Work-energy theorem.
According to the work-energy theorem, change in kinetic energy of the body is equal to the work done by the force.
$\therefore W = \dfrac{1}{2} m v_{f}^{2} - \dfrac{1}{2} m v_{i}^{2} $
$ W = \dfrac{1}{2} \times 0.5 \times (10\sqrt{2})^{2} - \dfrac{1}{2} \times 0.5 \times 0^{2} $
$\implies W = \dfrac{1}{2} \times 0.5 \times 200$
It gives,
$W = 50 J$
The work done by the net force during its displacement from $x=0$ to $x=2$ is $50 J$.

Note:Work produced by force is positive if the angle among force and displacement is acute. This implies, when the force and displacement are in the identical direction, work done is positive. When the force acts in a direction at normal to displacement, no work is done. Work produced by force is negative if the angle among force and displacement is obtuse. This implies, when the force and displacement are in reverse directions, work done is negative.