Question

A ball of mass 50 gm slows down from a velocity of 5m/sec to 3m/sec. Calculate the change in kinetic energy of the ball.

Answer 1

Since, we know velocity is one of the important factors on which kinetic energy depends, therefore, change in velocity leads to change in kinetic energy of the ball. Change in kinetic energy is the difference between final and initial KE where Final and Initial KE can be calculated with given mass and initial and final velocity accordingly.

Complete step by step answer:
Given that, mass of the ball = 50 gm. Now, velocity slows down because of the action of any resisting force from 5m/sec to 3m/sec.Initial velocity of ball = 5m/sec.Let ${{\text{v}}_1}$be the initial velocity of the ball. (We may also represent the initial velocity by u)
${{\text{v}}_1}$= 5m/sec
Also, Final velocity of ball = 3m/sec
Let ${{\text{v}}_2}$ be the final velocity of the ball.
\therefore $$$${{\text{v}}_2}= 3m/sec
Since, kinetic energy is the energy possessed by any object due to the virtue of its motion. Therefore, change in velocity of an object means the change in kinetic energy.The equation for calculating kinetic energy of any body is given as:
$KE = \dfrac{1}{2}m{v^2}$
Where, m = mass of body/object, V= velocity of the body/object

And the equation for calculating change in kinetic energy ($\Delta KE$) of any body is given as:
$\Rightarrow \Delta KE$= final kinetic energy – initial kinetic energy
$\Rightarrow \Delta KE = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2$
$\Rightarrow \Delta KE = \dfrac{1}{2}m(v_2^2 - v_1^2)$……………………………Eq.1
Substitute the numerical values in Eq.1

\Rightarrow \Delta KE = \dfrac{1}{2} \times \dfrac{{50}}{{1000}}({3^2} - {5^2}) \\\ \Rightarrow \Delta KE = 0.025 \times (9 - 25) \\\ \Rightarrow \Delta KE = 0.025 \times ( - 16) \\\ \Rightarrow \Delta KE = - 0.4{\text{J}} \\\ $$ Thus, the change in kinetic energy of the ball is -0.4 Joule (SI unit for any form of energy is Joule). Negative signs here represent the decreasing value of kinetic energy. Formula Used: $$ \Rightarrow KE = \dfrac{1}{2}m{v^2}$$ $$ \Rightarrow \Delta KE = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2$$ $$ \therefore\Delta KE = \dfrac{1}{2}m(v_2^2 - v_1^2)$$ **Note:** Remember that velocity was given in question with SI unit, thus we also need to convert the SI unit of mass, and else there would be an error in the resultant numerical value of kinetic energy. Also, do mention the significance of negative signs at the end of solution. There is a common misconception among students that kinetic energy can never be negative because mass can’t be negative and squared value of velocity is non-negative value. But it’s not the case, if the final speed is less than the initial speed of the object, then $$\Delta KE$$ is negative.