Asked by Jack
Driving in your car with a constant speed of v=13.0 m/s, you encounter a bump in the road that has a circular cross section. If the radius of curvature of the bump is r=33.2 m, calculate the force the car sear exerts on a 72.7 kg person as the car passes over the top of the bump.
No clue...
No clue...
Answers
Answered by
Count Iblis
When the car is over the top of the bump it is accelerating downward. The acceleration is
a = v^2/r = (13 m/s)^2/(33.2 m)
= 5.09 m/s^2
72.7 kg person is thus accelerated downward at this acceleration.
Newton's second law says that:
F = m a
Let'S apply this to the person. There are two forces that act on the person. There is the gravitational force exerted by the entire Earth on the person. This force is directed downward and has a magnitude of:
F_g = 72.7 kg * g = 713 N
And there is the foce exerted by the chair F_ch
If the take the downward direction to be positive, we have:
(F_g + F_ch) = 72.7 kg * 5.09 m/s^2 --->
F_ch = -343 N
The minus sign means that it is directed upward.
a = v^2/r = (13 m/s)^2/(33.2 m)
= 5.09 m/s^2
72.7 kg person is thus accelerated downward at this acceleration.
Newton's second law says that:
F = m a
Let'S apply this to the person. There are two forces that act on the person. There is the gravitational force exerted by the entire Earth on the person. This force is directed downward and has a magnitude of:
F_g = 72.7 kg * g = 713 N
And there is the foce exerted by the chair F_ch
If the take the downward direction to be positive, we have:
(F_g + F_ch) = 72.7 kg * 5.09 m/s^2 --->
F_ch = -343 N
The minus sign means that it is directed upward.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.