16) A curve of radius 120 m is banked at an angle of 18°. At what speed can it be negotiated under icy conditions where friction is negligible?

2 m/s 12 m /s 20 m /s 5 m/s 11 m/s

To find the speed at which the curve can be negotiated under icy conditions, we can use the formula for the critical speed at which friction is negligible:

v = √(rg tanθ)

Where:
v = speed
r = radius of the curve
g = acceleration due to gravity
θ = angle of banking

Given:
r = 120 m
θ = 18°
g = 9.8 m/s² (approximately)

Plugging in the values into the formula:

v = √(120 * 9.8 * tan18°)

Using a scientific calculator to calculate the value of tan18°:

tan18° ≈ 0.32492

v = √(120 * 9.8 * 0.32492)
v ≈ √(376.2)
v ≈ 19.39 m/s (approximately)

Therefore, the speed at which the curve can be negotiated under icy conditions where friction is negligible is approximately 19.39 m/s.

None of the given options (2 m/s, 12 m/s, 20 m/s, 5 m/s, 11 m/s) match the calculated value, so none of them is the correct answer.

To determine the speed at which the curve can be negotiated under icy conditions, we can use the concept of centripetal force and the equation for banked curves.

The centripetal force required to keep an object moving in a curved path is given by the equation:

F = mv^2/r

Where:
F - Centripetal force required
m - Mass of the object
v - Velocity of the object
r - Radius of the curve

In the case of the given question, the friction is negligible, so the centripetal force is provided by the horizontal component of the normal force. The normal force can be broken down into two components: the vertical component (Ncosθ) and the horizontal component (Nsinθ), where θ is the angle of inclination of the banked curve.

The horizontal component of the normal force provides the centripetal force:

F = Nsinθ = mv^2/r

Now, we can solve for v:

v = √(r * g * tanθ)

Where:
g - Acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the given values:
r = 120 m
θ = 18°

v = √(120 * 9.8 * tan(18°))
v ≈ √(120 * 9.8 * 0.3249)
v ≈ √(3781.44)
v ≈ 61.5 m/s

Since the options for speed provided are all in m/s, we can conclude that the correct answer is 12 m/s.

20m/s