# A 1075 kg car traveling 35 m/s puts on the brakes and skids to a stop. The coefficient of sliding friction between rubber tires and wet ?

pavement is 0.33. Show all forces acting on the car while it is skidding.

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• Given,

Mass (m) = 1075kg

Speed (v) = 35m/sec

Co-efficient of friction (μ) = 0.33

Frictional force (Ff) = ?

We know that:-

Frictional force (Ff) μN = μmg

= 0.33 x 1075 x 9.81

= 3480N

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• So, obviously, there is gravity downwards. Next, because the car isn't falling through the pavement, the pavement is exerting a normal force back up, which also is why there is friction. Lastly, there is the force of friction opposite to the direction of travel. So three forces in total.

Hope that helped

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F=ma so look for acceleration

F=mg where g is gravity

Ff=μ*NwhereF f is the frictional force (in Newtons),μ(mu) is the static or kinetic (μk) frictional coefficientN is the normal force (in Newtons). • Andrew Smith
Lv 7
1 month agoReport

Hi Jim, did you draw this yourself or find it elsewhere?  I draw the line at making drawings for others on line.  Too time consuming.

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