Question

A body of mass 5kg is moving with a momentum of 10kgm/s. A force of 0.2N acts on it in the direction of motion of the body for 10s. The increase in its kinetic energy is:
A.) 2.8 J
B.) 3.2 J
C.) 3.8 J
D.) 4.4 J

Answer 1

Hint: Change in momentum of the body on the application of an external force is equal to the product of force applied and the time for which the force acts on the body. By calculating the velocity from momentum, we can calculate change in kinetic energy.

Formula used:

Expression for momentum:

$p = m{\text{v}}$

Expression for kinetic energy:

$K = \dfrac{1}{2}m{{\text{v}}^2}$

Change in momentum: $\Delta p = F.t$

Detailed step by step solution:
We are given a body of mass 5kg.

$\therefore m = 5kg$

The momentum of the body is:

$p = 10kgm/s$

As we know that momentum is equal to the product of mass and velocity of the body, we can calculate the initial velocity as follows:

$\begin{gathered} p = m{\text{v}} \\\ {\text{v}} = \dfrac{p}{m} \\\ \end{gathered}$

Substituting the values of mass and momentum, we get

${\text{v}} = \dfrac{{10}}{5} = 2m/s$

This is the initial velocity of the body. Therefore the initial kinetic energy of the body is given as
$K = \dfrac{1}{2} \times 5 \times {\left( 2 \right)^2} = 10kg{m^2}{s^{ - 2}}$

Now an external force of 0.2 N acts on the body for 10s. Therefore the change in momentum of the body is given as

$\Delta p = F.t = 0.2 \times 10 = 2kgm/s \\\ \Delta p = p' - p = 2kgm/s \\\$

where p’ is the final momentum whose value can be calculated from this expression as follows:

$p' = p + 2 = 10 + 2 = 12kgm/s$

Therefore from the relation between momentum and velocity, we can calculate the final velocity as follows:

$p' = m{\text{v'}} \\\ \Rightarrow {\text{v'}} = \dfrac{{p'}}{m} = \dfrac{{12}}{5} = 2.4m/s \\\$

Hence the final kinetic energy of the body can be calculated now as follows:

$K' = \dfrac{1}{2}m{\text{v}}{{\text{'}}^2}$

Substituting the various values we get the final kinetic energy as follows:

$K' = \dfrac{1}{2} \times 5 \times {\left( {\dfrac{{12}}{5}} \right)^2} = 14.4kg{m^2}s$

Therefore, the increase in kinetic energy can be calculated the difference between final kinetic energy and initial kinetic energy as follows:

$\Delta K = K' - K = 14.4 - 10 = 4.4kg{m^2}{s^{ - 2}}$

Hence, the correct answer is option D.

Note: There is an increase in kinetic energy of the body because the force acts in the direction of motion of the body. If the force had been applied in the direction opposite to the direction of motion of the body then then there would have been a decrease in kinetic energy.