A car of mass 1800kg moving with a speed of 10m/s is brought to rest after a covering a distance of 50m. calculate the force acting on the car.
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Using third law of 1d kinematics,
v^2-u^2=2as
Given, s=50m; v=0; u=10m/s
Then , v^2-u^2=2as
0-(10)^2=2a(50)
a= -1m/s
Force=mass×acceleration, (or) Force=mass×deceleration
Therefore, deceleration= -1m/s^2
Deceleration= -(acceleration)
Force=mass×deceleration
=1800×1
=1800N
Force will be acting in opposite direction to displacement of the car
Hope it helps
Pls mark it as the brainliest.
v^2-u^2=2as
Given, s=50m; v=0; u=10m/s
Then , v^2-u^2=2as
0-(10)^2=2a(50)
a= -1m/s
Force=mass×acceleration, (or) Force=mass×deceleration
Therefore, deceleration= -1m/s^2
Deceleration= -(acceleration)
Force=mass×deceleration
=1800×1
=1800N
Force will be acting in opposite direction to displacement of the car
Hope it helps
Pls mark it as the brainliest.
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