Question

A cricket ball of mass 100g is moving with a velocity of 0.8m/s is stopped by a player in 0.5s. What is the force applied by the player to stop the ball?

Answer 1

Answer:

Magnitude of the force applied by the player is 0.16 N.

Step-by-step explanation:

Using first equation of motion :

• v = u + at, where symbols have their usual meaning.

Here,

Initial velocity u = 0.8 m/s

Final velocity v = 0

Time taken t = 0.5 s

= > 0 = ( 0.8 m/s ) + a( 0.5 s )

= > 0.8 / 0.5 m/s^2 = - a ( retardation )

= > 1.6 m/s^2 = - a

Force applied by player :

= > m a

= > 100 g x 1.6 m/s

= > 0.1 x 1.6 N

= > 0.16 N

Hence the magnitude of the force applied by the player is 0.16 N.

Answer 2

Answer:

Hope it help . Mark as brainliest please

Step-by-step explanation:

$f = ma \\ recall \: a \: = \frac{ {v} - {u}}{t} \\ since \: u = 0 \\ a = \: \frac{v}{t} \\ \\ f = m( \frac{v}{t}) \\ f = \:0.1kg( \frac{0.8}{0.5}) \\ f \: = 0.16 \: newton$