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Kryger [21]
2 years ago
14

A car moving at a steady 10 m/s on a level highway encounters a bump that has a circular cross-section with a radius of 30 m. Th

e car maintains its speed over the bump. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?
Physics
1 answer:
daser333 [38]2 years ago
8 0

When you are driving over a circular bump, you can feel your body moving up a small distance in the seat. This make you feel like you weigh less that you really do. This is caused by the fact that net down force on you is equal to the centripetal force. If you go the following website, you will see a problem that is similar to this one. Go the problem called Sample Roller Coaster Problem.

At this website you can see that the normal force exerted by the seat of the car on a 60.0-kg passenger is equal to the difference of the passenger’s weight and the centripetal force.

Weight = 60 * 9.8 = 588 N

Fc = 60 * 10^2/30 = 200 N

Normal force = 388 N

Explanation:

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As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

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Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

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