Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved-that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.)

(a) On dry concrete.

(b) On wet concrete.

(c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

a) a = - 6.28 m/s²

b) a = - 4.08 m/s²

c) a = - 1.69 m/s²

Explanation:

Given that

θ = 4º

When car is going up then friction force will act downwards

Lets take a is the acceleration a m/s²

From second law of Newton's

ma = - mg sin θ - μ mg cosθ

a = - g sin θ - μ g cosθ

a)

For dry concrete

μ = 0.7

a = - g sin θ - μ g cosθ

a = - 10 (sin4º - 0.7 cos4º)

a = - 6.28 m/s²

b)

For wet concrete

μ = 0.34

a = - g sin θ - μ g cosθ

a = - 10 (sin4º - 0.34 cos4º)

a = - 4.08 m/s²

c)

For ice

μ = 0.1

a = - g sin θ - μ g cosθ

a = - 10 (sin4º - 0.1 cos4º)

a = - 1.69 m/s²