Question

Certain force acting on a 20 kg mass changes its velocity from $5m/s$ to $2m/s$ . Calculate the work done by the force.

Answer 1

Hint: Work energy theorem gives us that work done in a system is given by the change in kinetic energy. Find the initial and final kinetic energy of the system with the given values. Find the difference between them to find the work done.

Complete step by step answer:
The principle of work energy theorem states that the work on a system by the forces acting on it is equal to the change in kinetic energy of the system.
We can simply write it as the difference in initial and final kinetic energy of the system is the work done on the system.
Mathematically we can write,
$W=\Delta KE=K{{E}_{f}}-K{{E}_{i}}$
Given in the question that,
The mass of the particle is $m=20kg$
Initial velocity of the particle is ${{v}_{i}}=5m/s$
Final velocity of the particle is ${{v}_{f}}=2m/s$
Initial kinetic energy of the body is
$\begin{aligned} & K{{E}_{i}}=\dfrac{1}{2}mv_{i}^{2} \\\ & K{{E}_{i}}=\dfrac{1}{2}\times 20\times {{5}^{2}} \\\ & K{{E}_{i}}=250\text{ joule} \\\ \end{aligned}$
Again, final kinetic energy of the body is
$\begin{aligned} & K{{E}_{f}}=\dfrac{1}{2}mv_{f}^{2} \\\ & K{{E}_{f}}=\dfrac{1}{2}\times 20\times {{2}^{2}} \\\ & K{{E}_{f}}=40\text{ joule} \\\ \end{aligned}$
So, change in kinetic energy of the object is given as,
$\begin{aligned} & \Delta KE=K{{E}_{f}}-K{{E}_{i}} \\\ & ~\Delta KE=40J-250J \\\ & \Delta KE=-210J \\\ \end{aligned}$
Again, according to the work energy theorem, the work done on a system by a force is given by the change in kinetic energy of the system.
So, the work done is
$W=\Delta KE=-210J$
So, the work done on the system is -210 J.

Note: The work energy theorem is also known as the principle of work and kinetic energy which states that the total work done on a system by all the forces is equal to the change in kinetic energy of the system.
Kinetic energy is the energy obtained by a system in motion. For a system at rest, the kinetic energy is zero.