Question

A ball of mass 200 gram moving with a velocity of 10 metre per second is stopped by a boy in 0.2 second calculate the force applied by the boy to stop the ball​

Answer 1

Explanation:

Mass of ball=200g= 0.2kg

Velocity of ball=10m/s

Ball stop by boy in 0.2sec.

Force=mass×acceleration

but acceleration=(initial velocity - final velocity) /time

but initial velocity=10m/s and final velocity=0m/s

Time=0.2second

Acceleration=

(10−0) /0.2

=50 m/s2

Force=m×a=0.2×50=10N

Answer 2

Answer:-

• 10 newton force applied by boy to stop the ball.

Given:-

• Mass of ball ,m = 200g
• initial velocity ,u = 10m/s
• Final velocity ,v = 0m/s
• Time taken ,t = 0.2s

To Find:-

• Force applied to stop the ball ,F

Solution:-

According to the Question

It is given that ball of mass 200 gram moving with a velocity of 10 metre per second is stopped by a boy in 0.2 second. We have to calculate the force applied to stop the ball.

Firstly we calculate the acceleration of the ball.

As we know that acceleration is defined as the rate of change in velocity.

• 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 = 0-10/0.2

→ a = -10/0.2

→ a = -100/2

→ a = -50m/s².

Now, calculating the force applied by the boy to stop the ball.

Force is calculated by the product of mass & acceleration .

• F = ma

where,

• F denote Force
• m denote mass
• a denote acceleration

Substitute the value we get

→ F = 0.2 × (-50)

→ F = 2 × (-5)

→ F = -10 N

Here, negative sign show that the force is acting in opposite direction of motion .

• Hence, the force applied by boy to stop the ball is 10 Newton .