Question

A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force applied by the player to stop the ball?​

Answer 1

Answer:

Force applied is -0.07 N.

Explanation:

To find : Force applied by the player to stop the ball.

Given that,

• Mass of ball, m is 70 g.
• Initial velocity, u is 0.5 m/s.
• Final velocity, v is 0m/s.
• Time, t is 0.5 s.

We know,

• Force = ma

[Where, m is mass and a is Acceleration]

Acceleration = v - u/t

→ Acceleration = 0 - 0.5/0.5

→ Acceleration = -0.5/0.5

→ Acceleration = -1

Acceleration is -1 m/s² .

Conversion :

• m = 70 g = 70/1000 kg = 0.07 kg

Mass is 0.07 kg .

Now, Put m and a in force formula :

→ Force = 0.07 × -1

→ Force = -0.07

Thus,

Force applied by the player to stop the ball is - 0.07 N.

Answer 2

Given :-

A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s

To Find :-

Force applied

Solution :-

We know that

1 kg = 1000 g

70 g = 70/1000 = 0.07 kg

Now

F = m(v - u)/t

F = 0.07(0 - 0.5/0.5)

F = 0.07(-0.5/0.5)

F = 0.07(-1)

F = -0.07 N

Note - Negative sign shows that force is applied in the opposite direction

Hence,

Force applied is -0.07 N

Know More :-

Force - A push or pull applied on an object is called force. SI unit of Force is Newton(N)

Acceleration - The change of velocity with respect to time is called Acceleration.  SI unit of Acceleration is meter per second (m/s²)

Mass - The Weight of an object is called mass.  SI unit of mass is  Kilogram(Kg)