A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car when the car is at the very top of the hill?

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A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at
a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car
when the car is at the very top of the hill?
IS
Transcribed Image Text:A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car when the car is at the very top of the hill? IS
Mass of the car
(m)
4540 kg
- The radius of curvature of the hill (r) = 13.9.
-Speed of the
car v = 8.5 m/s
the magnitude of the normal force
acting
Given data
Finding
At the top of the
Fnet
-N
mg
-N = my ²
mg
mg
(mye is the centripetal force in circular motion, mg is
the
weight
my²
OR
the hill, the netforce
²² = N
N =
mg
454
2
= 4540 kg x 9.8 m/s² =
(9.8 m/s² -
= 4540 kg
my2
on the car,
4540 kg (8.5m/s)?
(8.5 m/s2)
13.9m
2
- 2
(9.8. m/s² - 5. 197 ms.
= 4540 kg
= 4540 kg x 4.603 m/s²
=20897.62 N
Transcribed Image Text:Mass of the car (m) 4540 kg - The radius of curvature of the hill (r) = 13.9. -Speed of the car v = 8.5 m/s the magnitude of the normal force acting Given data Finding At the top of the Fnet -N mg -N = my ² mg mg (mye is the centripetal force in circular motion, mg is the weight my² OR the hill, the netforce ²² = N N = mg 454 2 = 4540 kg x 9.8 m/s² = (9.8 m/s² - = 4540 kg my2 on the car, 4540 kg (8.5m/s)? (8.5 m/s2) 13.9m 2 - 2 (9.8. m/s² - 5. 197 ms. = 4540 kg = 4540 kg x 4.603 m/s² =20897.62 N
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