When a car of mass 1200 kg is moving with a velocity of 15m/s on a rough horizontal surface?
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A car with a mass of 1,000 kg is moving with a velocity of 20 m/s. If on applying brakes, the car stops after coming to a distance of 50 m, what will be its retarding force?
The car stops after traveling 50m. So, from the laws of motion,
V^2=U^2+2D(-a)
where V is the velocity at the end
U us its initial velocity
D is the distance traveled before coming to rest after applying retarding force
a us the acceleration but negative because retarding.
So
0=20^2–2*50*a
Or, 100a=400
So, a=4 ms^-2
Since the mass of the car is 1000 kg. So
F=m*a
=1000*4
=4000N
So the applied force is 4000 Newton's
this is just a example
I hope this will help you
The car stops after traveling 50m. So, from the laws of motion,
V^2=U^2+2D(-a)
where V is the velocity at the end
U us its initial velocity
D is the distance traveled before coming to rest after applying retarding force
a us the acceleration but negative because retarding.
So
0=20^2–2*50*a
Or, 100a=400
So, a=4 ms^-2
Since the mass of the car is 1000 kg. So
F=m*a
=1000*4
=4000N
So the applied force is 4000 Newton's
this is just a example
I hope this will help you
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