Question

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

Answer 1

Answer:

• 0.07 Newton is the required answer.

Explanation:

Given:-

• Mass ,m = 70g = 70/1000 = 0.07kg
• Initial velocity ,u = 0.5m/s
• Final velocity ,v = 0m/s
• Time taken ,t = 0.5 s

To Find:-

• Force ,F

Solution:-

Firstly we calculate the acceleration of the ball. As we know that Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

→ a = 0-0.5/0.5

→ a = -0.5/0.5

→ a = -1 m/s²

Here, negative sign show retardation

Acceleration of the ball is 1 m/s².

Now, Calculating the force applied by the player to stop the ball .

Force is the product of mass and acceleration.

• F = ma

Substitute the value we are

→ F = 0.07 × 1

→ F = 0.07 N

• Hence, the force required to stop the ball is 0.07 Newton.