Question

A cricket ball of mass 150g moving at a speed of 25m/s is brought to rest by a player in 0.03 seconds. find the average force applied by the player.​

Answer 1

Answer:124.99 N

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

Answer:

The average force applied by the player is 125N.

Explanation:

The force applied is given as,

$F=m\times a$          (1)

Where,

F=force applied to the body

m=mass of the body

a=acceleration of the body

Now we will use the first equation of motion to find acceleration,

$v=u+at$           (2)

v=final velocity=0     (body brought to rest)

u=initial velocity=25m/s (given)

a=acceleration

t=time taken=0.03 seconds(given)

So,

$0=25+a\times 0.03$

$a=\frac{2500}{3} ms^-^2$                (3)

The mass of the cricket ball given in the question is 150 g i.e. 0.15kg

By substituting the required values in equation (1) we get;

$F=0.15\times \frac{2500}{3}=125N$

Hence, the average force applied by the player is 125N.