Question

QUESTION 3 (8 marks) 1 40 ft 3 $\rho_2 = 20$ ft 2 15 ft Using 1 ft = 0.31 metres, for the (180+YZ/4)-kg amusement car: Find the velocity at position 2 and at position 3 (it starts from rest at 1) (2) Also find the normal force exerted by the surface on the car at 2. (4) What is the radius of curvature at 3 so the car doesn't go flying off the road? (hint-normal force from the surface is zero) (2)

          QUESTION 3 (8 marks)

1
40 ft
3
$\rho_2 = 20$ ft
2
15 ft
Using 1 ft = 0.31 metres, for the
(180+YZ/4)-kg amusement car:
Find the velocity at position 2 and
at position 3 (it starts from rest at
1) (2)
Also find the normal force exerted
by the surface on the car at 2. (4)
What is the radius of curvature at
3 so the car doesn't go flying off
the road? (hint-normal force from
the surface is zero) (2)

QUESTION 3 (8 marks)

1
40 ft
3
ρ2 = 20 ft
2
15 ft
Using 1 ft = 0.31 metres, for the
(180+YZ/4)-kg amusement car:
Find the velocity at position 2 and
at position 3 (it starts from rest at
1) (2)
Also find the normal force exerted
by the surface on the car at 2. (4)
What is the radius of curvature at
3 so the car doesn't go flying off
the road? (hint-normal force from
the surface is zero) (2)

Added by Samuel P.

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition

Chapter 5

Instant Answer

Solved by Expert Narayan Hari

Step 1

First, let's convert the given distances from feet to meters using the conversion factor 1 ft = 0.31 meters. - 40 ft = 40 * 0.31 = 12.4 meters - 20 ft = 20 * 0.31 = 6.2 meters - 15 ft = 15 * 0.31 = 4.65 meters Show more…

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QUESTION 3 (8 marks) Using 1 ft = 0.31 meters, for the (180 + YZ/4)-kg amusement car: Find the velocity at position 2 and at position 3 (it starts from rest at 1) (2) 40 ft = 20 ft 15 ft Also find the normal force exerted by the surface on the car at 2. (4) 2 What is the radius of curvature at 3 so the car doesn't go flying off the road? (hint: normal force from the surface is zero) (2)