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a car of mass 1000 kg slams on its brakes. the wheels lock and the tires skid on dry pavement. what force of friction is exerted on the car of hte coefficient of friction between the tires and the road

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Answer:

The force of friction exerted on the car when the wheels lock and the tires skid on dry pavement is 7840 N with the coefficient of friction between the tires and the road is 0.8

Step-by-step explanation:

The force of friction exerted on a car when the wheels lock and the tires skid on dry pavement is determined by the coefficient of friction between the tires and the road.

To calculate the force of friction, we can use the formula:

force of friction = coefficient of friction * normal force

The normal force is the force exerted on an object perpendicular to the surface it is in contact with. In this case, the normal force is equal to the weight of the car, which can be calculated using the formula:

weight = mass * gravity

where mass is in kg and gravity is 9.8 m/s^2 on earth

So the force of friction can be calculated as follows:

force of friction = coefficient of friction * (mass * gravity)

if the coefficient of friction between the tires and the road is 0.8

force of friction = 0.8 * (1000 kg * 9.8 m/s^2)

force of friction = 7840 N

Therefore, the force of friction exerted on the car when the wheels lock and the tires skid on dry pavement is 7840 N with the coefficient of friction between the tires and the road is 0.8

User Fedor Hajdu
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