Question

The speed of a ball of mass 300 g increases from 5m/s to 7m/s. What is the increase in kinetic

energy of the ball?​

Answer 1

Explanation:

Given that:-

Mass of ball =300g=0.3 kg

Speed v1=5 m/s

Speed v2=7 m/s

Increase in kinetic energy

=KE2-KE1

$\frac{1}{2} mv2 {}^{2} - \frac{1}{2} mv1 {}^{2} \\ \frac{1}{2} m(v2 {}^{2} - v1 {}^{2} ) \\ \frac{1}{2} \times 0.3(7 {}^{2} - 5 {}^{2} ) \\ \frac{0.3}{2} \times (49 - 25) \\ \frac{0.3}{2} \times 24 \\ 0.3 \times 12 \\ 3.6$

Therefore, increase in kinetic energy is 3.6 J

Answer 2

$\huge\mathfrak\red{ANSWER}$

INCREASES BY 3.6J

Explanation:

INCREASE IN KINETIC ENERGY OF BALL

= FINAL KINETIC ENERGY - INITIAL KINETIC ENERGY.

KINETIC ENERGY IS ENERGY OF BODY DUE TO IT'S MOTION. IT IS MEASURED IN JOULES.

FOR A BODY OF MASS (M) MOVING WITH VELOCITY (V) KINETIC ENERGY OF THIS BODY IS

$k.e = 1 \div 2 \times m {v}^{2}$

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NOW COMING TO QUESTION------>

KINETIC ENERGY OF BALL WHEN VELOCITY WAS 5

= 1/2 × 0.3kg×5^2

=0.15×25

=3.75 joules

KINETIC ENERGY OF BALL WHEN VELOCITY WAS 7

=1/2 × 0.3kg × 7^2

=0.15 × 49

=7.35 joules.

THUS CHANGE IN KINETIC Energy

= 7.35J - 3.75J

= 3.6J

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BY SCIVIBHANSHU

THANK YOU

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