A 1500 kg car sits on a 3.5° inclined hill. Find the force of friction required to keep it from sliding down the hill. The coefficient of static friction is μ=0.45
The same 1500 kg car is coasting at 50 km/h when it encounters a (friction
free!) hill that drops 14 m vertically. It then travels through 30.0 m of mud with an effective kinetic friction coefficient of 0.25. Determine the speed of the car after it emerges from the mud (in km/h).

a. 6602.7 N b. 64.44 km/h

Explanation:

a. Find the force of friction required to keep it from  sliding down the hill.

The frictional force f equals the component of the weight of the car, W perpendicular to the inclined hill = Wcosθ times the coefficient of static friction, μ = 0.45.

Since f = μN = μWcosθ = μmgcosθ where m = mass of car = 1500 kg, g = acceleration due to gravity = 9.8 m/s² and θ = angle of incline of hill = 3.5°

So, f = μmgcosθ

= 0.45 × 1500 kg × 9.8 m/s²cos3.5°

= 6615cos3.5°

= 6602.7 N

b. Determine the speed of the car after it emerges from the mud (in km/h)

Since the car drops a vertical height of 14 m, its potential energy decreases by mgh and its kinetic energy increases by mgh where m =mass of car and h = height drop = 14 m. So its kinetic energy increase is ΔK = mgh = 1500 kg × 9.8 m/s² × 14 m = 205800 J

Since it has an initial velocity of u = 50 km/h = 50 km/h 1000m/3600 s = 13.89 m/s, its initial kinetic energy is K = 1/2mu² = 1/2 × 1500 kg × (13.89 m/s)² = 144699.08 J.

Its new kinetic energy after the drop is thus K' = K + ΔK = 144699.08 J + 205800 J = 350499.08 J

Let v be its velocity after the drop, since K' = 1/2mv²,

v = √(2K'/m) = √(2 × 350499.08 J/1500 kg) = √(700998.16 J/1500 kg) = √(467.332 J/kg) = 21.62 m/s

Now, from work kinetic energy principles, the kinetic energy change in the car is the work done on car by friction

So, ΔK" = -fd = -μmgd

Let v' be the velocity of the car after emerging from the mud and moving a distance d = 30.0 m.

So, 1/2m(v'² - v²) = -μmgd

v'² - v² = -2μgd

v'² = v² - 2μgd

Substituting the values of the variables, we have

v'² = (21.62 m/s)² - 2 × 0.25 × 9.8 m/s² × 30.0 m

v'² = 467.42 m²/s² - 147 m²/s²

v'² = 320.42 m²/s²

taking square root of both sides, we have

v' = √(320.42 m²/s²)

= 17.9 m/s

Converting v to km/h we have v' = 17.9 m/s × 3600 s/h × 1 km/1000 m = 64.44 km/h.

So, the car emerges from the mud with a speed of 64.44 km/h

d is the answer i came up with