Question

a 2000kgvehicle is moving with a speed of 20ms^-1 brought to rest in a distance of 50mby applying brakes.What is the acceleration and magnitude of net force acting on body

Answer 1

u = 20 m/s
v=0
mass = 2000 Kg
s= 50 m

=> ( v )^2 - ( u )^2 = 2 a s

0 - 400 = 2 . a .50

-400 = 100 a

a = - 4 m/s^2

force = mass × acc.
force = 2000 × -4
force = -8000 Newton

Answer 2

$\red{ HEY\: BUDDY !!! }$

HERE'S THE ANSWER...

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▶️ First we'll note the following data

⏺️ initial velocity ( u ) = 20 m/s

⏺️final velocity ( v ) = 0 { going to rest }

⏺️Mass ( m ) = 2000 Kg

⏺️ displacement ( S ) = 50 m

⏺️ Acceleration ( a ) = ?

⏩ According to third equation of motion

⚫ ( v )^2 - ( u )^2 = 2 × a × S

=> ( 0 )^2 - ( 20 )^2 = 2 × 50 × a

=> - 400 = 100 × a

=> $\boxed{ a\:=\: -4 \:m/s^2 }$

↪️ Here acceleration is in negative because velocity is decreasing.

⏩ Now we know , Force is product of mass ( m ) and acceleration ( a ) , i.e

⚫ Force ( F ) = mass ( m ) × acceleration ( a )

=> F = 2000 × ( - 4 )

=> $\boxed{ F\:=\: -8000 \:Newton ( N ) }$

↪️ Here force is in negative because it is in opposite direction of motion..

HOPE HELPED...

$\red{ JAI\:HIND..}$

:)