Calculate the maximum acceleration (in m/s 2 ) of a car that is heading up a 14.

Question

Calculate the maximum acceleration (in m/s2) of a car that is heading up a 14.5° slope (one that makes an angle of 14.5° 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 static coefficient of friction is involved--that is, the tires are not allowed to slip during the acceleration.

(a)

on dry concrete

m/s2

(b)

on wet concrete

m/s2

(c)

on ice, assuming that

µs = 0.100,

the same as for shoes on ice

m/s2

Explanation / Answer

The friction force must be sufficient to overcome the component of the weight that is parallel to the ramp and accelerate the car.

Friction force - Parallel force = ma

s (0.5 m) g cos(14.5°) - m g sin(14.5°) = ma

Note: We use (0.5 m) in the above equation because only half of the car's weight contributes to the required friction force. s is used because we are not skidding. k is extraneous information and is not required.

Cancel m's.

a = s (0.5) g cos(14.5°) - g sin(14.5°)

= g[(0.5* s* cos(14.5°)) – (sin(14.5°))]

Part (a)

s = 1

a =9.8 [(0.5* 1* cos(14.5°)) – (sin(14.5°))]

= 2.29 m/s2

Part (b)

s = 0.7, pugging in we get

a = 0.867 m/s2

Part (c)

s = 0.1, pugging in we get

a = -2.01 m/s2

Hope this helps :)

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