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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?

Respuesta :

Answer:

(a) 388 newton

(b) 17.15 m/s

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

Answer:

(a) Normal force will be 388 N

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

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 [tex]N=mg-\frac{mv^2}{r}=60\times 9.8-\frac{60\times 10^2}{30}=588-200=388N[/tex]

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

So [tex]0=mg-\frac{mv^2}{r}[/tex]

[tex]mg=\frac{mv^2}{r}[/tex]

[tex]60\times 9.8=\frac{60\times v^2}{30}[/tex]

[tex]v^2=294[/tex]

v = 17.94 m/sec