If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N , what is the normal force on the car when it is at the bottom of the track (point A

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Answer:

So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.

Now,

We have F=ma,

and a=v/t,

so we can have another equation for it as,

Now, providing the required data  to, it;  ∴t =2 sec,

F=(1)×(20/2),

  • F=10 N.
  • So, the car would be acting the force,F of about 10 N  while the car is present on the lower region of the track.