Physics, asked by bhandarikri321, 6 months ago

a cricket ball of mass 150 g moving at a speed of 25 mls is brought to rest by a player in 0.03s . Find the average force applied by the player .​

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
10

Answer

  • Mass of the ball = 150 g
  • Initial Velocity = 25 m/s
  • Final Velocity = 0 m/s
  • Time = 0.03 sec
  • Average Force = ?

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  • First Find the acceleration of the body with the help of the first equation of motion as we are given the initial and the final Velocity along with the time taken. Then find the Force Applied using F = ma (Newton's Second Law)

\sf :\implies v = u+at

\sf :\implies 0 = 25+ a\times 0.03

\sf :\implies -25 = 0.03t

\sf :\implies \dfrac{-25}{0.03} = a

\sf :\implies\underline{\boxed{\red{\mathfrak{acceleration = 833.3 \ m/s^2}}}}

  • Now the Force Applied which will be equal to the product of the acceleration and the mass of the body. But here we are given the mass in grams, so we have to convert it to kg
  • 150 g = 150/1000 = 0.15 kg

\sf :\implies F = ma

\sf :\implies F = 0.15 \times 833.3

:\implies\underline{\boxed{\pink{\mathfrak{f \approx 125} \ \sf N}}}

\displaystyle\therefore\:\underline{\textsf{ The Force Applied is \textbf{125 N}}}


prince5132: Nice !
Answered by VinCus
100

\rule{220}{1}

{ \huge{ \underline{ \underline{ \underline{ \frak{ \red{Required \: Answer : }}}}}}}

Given:

--Mass of ball (m) = 150g =15/100kg

-- Initial velocity (u) 25m/s

-- Final velocity (v) = Om/s

--Time taken (t) = 0.03s

\rule{220}{1}

To Find:

--Average Force applied on it (F)

\rule{220}{1}

Solution:

--As we know Force is the product of mass and acceleration.

--Firstly we have to calculate the acceleration of the ball. So by using 1st equation of motion

 \sf \boxed{v = u + at}

--Substitute the value we get

--0 = 25 + a × 0.03

---25 = a × 0.03

-- a=-25/0.03

-- a= -833.33 m/s²

--here negative sign show retardation

--The acceleration of the ball is 833.33m/s²

--Now using this Formula

 \sf \boxed{F = m \times a}

--Substitute the value we get

--F= 15/100x 833.33

--F = 0.15 x833.33

--F= 124.9 = 125N

 \huge \sf \boxed{125 \: N}

Hence Proved.

\rule{220}{1}

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