A 1600 kg car rounds a curve of 61 m banked at an angle of 18 ° . If the car is traveling at 97 km/h , will a friction force be required? If so, in what direction? How much friction force will be required?

Explanation:

With no friction, the normal force of slope on car will provide the necessary centripetal force in the horizontal and support the weight of the car in the vertical

tanθ = mv²/R / mg = v²/Rg

97 kph = 26.94 m/s

tanθ = 26.94²/(61(9.8)) = 1.214458...

θ = 50.5°

As this is much steeper than the 18° bank angle, a friction force is required in the down slope direction. or toward the center of curvature.

The ideal speed for an 18° slope is

v = √(Rgtanθ) = √(61(9.8)tan18) = 13.93 m/s

The 18° slope at ideal speed makes a centripetal force of

Fc = 1600(13.93²/61) = 5,094.7

The required centripetal force at 97 kph is

Fh = 1600(26.94²) / 61 = 19,042.7 N

Tire friction must supply 19042.7 - 5094.7 = 13,948 N horizontally

Therefore friction parallel to the slope must be

13,948 / cos18 = 14,665 N = 15 kN

t = 1.7 x 10⁻⁷ s

Explanation:

Given,

The mass of the electron, m = 9.1 x 10⁻³¹ Kg

The force experienced by the electron, F = 1.6 x 10⁻¹⁷ N

The initial velocity of the electron, u = 0

The final velocity of electron, v = 3.0 x 10⁶ m/s

The acceleration of the electron is given by the relation

                                         a = F/m

                                            = 1.6 x 10⁻¹⁷ N / 9.1 x 10⁻³¹ Kg

                                            = 1.76 x 10¹³ m/s²

Using the first equation of motion

                                        v = u + at

                                        t = (v - u) / a

                                          = (3.0 x 10⁶ - 0) /  1.76 x 10¹³

                                          = 1.7 x 10⁻⁷ s

Hence, the time taken by the electron to reach final speed is, t = 1.7 x 10⁻⁷ s