A motorcycle is moving with a velocity of 90 km/h and it takes 5 seconds to stop after the breaks are applied.calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 200 kg.
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8
Convert 90 km/h to m/s
=> (90 x 5/18) m/s
=> 25 m/s
Now,
=> V = u+at
=> 0 = 25 + a x 5
=> -25 = a x 5
=> -25/5 = a
=> a = -5 m/s^2
Now,
Force = mass x Acceleration
=> F = ma
=> F = {200 x (-5)} Kg m/s^2
=> F = -1000 Kg m/s^2
=> F = -1000 N.
Be Brainly
=> (90 x 5/18) m/s
=> 25 m/s
Now,
=> V = u+at
=> 0 = 25 + a x 5
=> -25 = a x 5
=> -25/5 = a
=> a = -5 m/s^2
Now,
Force = mass x Acceleration
=> F = ma
=> F = {200 x (-5)} Kg m/s^2
=> F = -1000 Kg m/s^2
=> F = -1000 N.
Be Brainly
Answered by
3
mass of motorcycle along with the rider , M = 200g
initial velocity of motorcycle, u = 90km/h = 90 × 5/18 = 25 m/s
a/c to question,
motorcycle takes 5 sec to stop after the breaks are applied.
so, final velocity of motorcycle, v = 0
use formula, v = u + at
0 = 25 + a × 5 => a = -5 m/s²
hence, retardation applied on motorcycle is 5m/s² .
now, use Newton's 2nd law,
e.g., F = Ma
M = 200kg, and a = 5m/s²
hence, F = 200 × 5 = 1000N
hence, 1000N force exerted by the brakes on the motorcycle
initial velocity of motorcycle, u = 90km/h = 90 × 5/18 = 25 m/s
a/c to question,
motorcycle takes 5 sec to stop after the breaks are applied.
so, final velocity of motorcycle, v = 0
use formula, v = u + at
0 = 25 + a × 5 => a = -5 m/s²
hence, retardation applied on motorcycle is 5m/s² .
now, use Newton's 2nd law,
e.g., F = Ma
M = 200kg, and a = 5m/s²
hence, F = 200 × 5 = 1000N
hence, 1000N force exerted by the brakes on the motorcycle
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