Question

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. The car maintains its speed over the bump.What is the normal force exerted by the seat of the car on a 60 kg passenger when the car is at the top of the bump? Part B) what speed does the car lose contact with the bump road?

Answer 1

(a) Normal force will be 388 N

(B) Speed of the car will be 17.94 m /sec

Step-by-step explanation:

We have given velocity of the car v = 10 m /sec

radius of the bump = 30 m

Mass of the passenger m = 60 kg

(a) When the car is at the top of the bump then normal force will be equal to weight of the body minus centripetal force

So normal force
`N=mg-(mv^2)/(r)=60* 9.8-(60* 10^2)/(30)=588-200=388N`

(B) When car loose the contact then normal force will be zero

So
`0=mg-(mv^2)/(r)`

`mg=(mv^2)/(r)`

`60* 9.8=(60* v^2)/(30)`

`v^2=294`

v = 17.94 m/sec

Answer 2

(a) 388 newton

(b) 17.15 m/s

Step-by-step explanation:

r = 30 m

v = 10 m/s

m = 60 kg

(a) Let N be the normal reaction.

At the bump

N = mg - mv^2 / r

N = 60 x 9.8 - 60 x 10 x 10 / 30 = 588 - 200 = 388 newton

(b) Then the contact loose, N = 0

So, mg = mv^2 / r

v^2 = r x g = 30 x 9.8 = 294

v = 17.15 m/s