Question

A cricket ball of 100 g moving at a speed of 40 ms^-1 is to brought to rest by a player in 0.04 sec. Find the force
applied by the player?

Plz anyone solve this numerical......​

Answer 1

Answer- The above question is from the chapter 'Force and Laws of Motion'.

Concept used: 1) v = u + at (Ist Equation of Motion)

v - u = at

$\frac{v \: - \: u}{t} = a$

Here, v = final velocity

u = initial velocity

t = time

a = acceleration

2) F = ma (Newton's Second Law of Motion)

Here, m = mass of body

F = Force applied

Given question: A cricket ball of 100 g moving at a speed of 40 m/s is to brought to rest by a player in 0.04 s. Find the force applied by the player.

Answer: Mass of ball = 100 g = 0.1 Kg

Initial speed (u) = 40 m/s

Final speed (v) = 0 m/s

Time (t) = 0.04 s

Acceleration (a) = $\frac{v \: - \: u}{t}$

a = $\frac{0 \: - \: 40}{0.04}$

a = - 1000 m/s²

F = ma

F = 0.1 × - 1000

F = - 100 N

∴ Force applied by the player = 100 N in the direction opposite to the ball thrown.

Answer 2

SHORT / DIRECT METHOD :-

$F = \frac{mdv} {dt}$

$F = 100^{-3} X \frac{40^{10}}{4 X 10^{-2}}$

$F = 100 X \frac{10^{-2}}{ 10^{-2}} = 100 N$

LONG METHOD :-

Concept used: 1) v = u + at (Ist Equation of Motion)

v - u = at

Here, v = final velocity

u = initial velocity

t = time  

a = acceleration

 

2) F = ma (Newton's Second Law of Motion)

Here, m = mass of body  

F = Force applied

 

Mass of ball = 100 g = 0.1 Kg

Initial speed (u) = 40 m/s

Final speed (v) = 0 m/s

Time (t) = 0.04 s

Acceleration (a) =  

a =  

a = - 1000 m/s²  

F = ma

F = 0.1 × - 1000

F = - 100 N  

∴ Force applied by the player = 100 N in the direction opposite to the ball thrown.

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