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

Given:
mass of  ball =m=70g=0.070kg
Initial velocity=u= 0.5 m/s
Final velocity =v=0m/s
time=t= 0.5 sec
a= v-u/t
⇒0-0.5/0.5
⇒-1 m/s²
F=m*a
⇒0.07x-1
⇒-0.07 N

Answer 2

The force is $-0.07N$

Given:

Mass of the cricket ball= 70 g or 0.07 kg

The initial velocity of ball (u)= 0.5 $\frac{m}{s}$

To find:

The force applied

Solution:

We are given that

Mass of ball= 70 g or 0.07 kg

Initial velocity of ball (u)= 0.5 $\frac{m}{s}$

The final velocity (v)= 0 $\frac{m}{s}$

Final velocity is 0 because the ball is stopped.  

Time taken (t)=0.5 s

We now calculate the force applied to stop the ball.

This can be calculated using the formula:

$\begin{aligned} & F=m \times\left[\frac{v-u}{t}\right] \\\\=& 0.07 \times[(0-0.5) / 0.5] \\\\=& 0.07 \times(-) 1=-0.07 N \end{aligned}$

In this case, the negative sign shows that the force is a retarding force.