Question

10. A cricket ball of mass 150 g moving at a speed of
25 ms' is brought to rest by a player in
003 s. Find the average force applied by the
player​

Answer 1

Answer:

124.99N

Explanation:

m=150g=0.15kg

v=25m/s

t=o.03s

F=ma

a=v/t

=25/0.03

=833.33m/s^2

therefore

F=ma

=0.15*833.33

=124.99N.

Answer 2

GIVEN :-

• Mass of cricket ball , m = 150 g = 0.15 kg.
• Initial velocity , u = 25 m/s.
• Final velocity , v = 0 m/s. [brought to rest]
• Time , t = 0.03 seconds.

TO FIND :-

• The average force applied by the player , F.

SOLUTION :-

We will find the acceleration by using the first equation of motion.

★ First Equation of motion ★

• v = u + at

➳ 0 m/s = 25 m/s + a × 0.03s

➳ 0 m/s - 25 m/s = a × 0.03s

➳ -25 m/s = a × 0.03s

➳ a = (-25 m/s)/0.03s

➳ a = -25/0.03 m/s².

[ Note :- The -ve sign in acceleration shows the retardation. ]

★ By using the formula of Force ★

• F = ma

➠ F = 0.15 kg × -25/0.03 m/s²

➠ F = 5 kg × -25 m/s²

➠ F = -125 kg . m/s²

➠ F = -125 N.

Hence the average force applied by the player is 125 Newtons.