# Billy tests the acceleration of his 950kg automobile, and finds that it can go 0 to 100km/h (27.8m/s) in 8.52 seconds. The engine...provides a force of 10.74kN (200hp), what is the average force of...

Billy tests the acceleration of his 950kg automobile, and finds that it can go 0 to 100km/h (27.8m/s) in 8.52 seconds. The engine...

provides a force of 10.74kN (200hp), what is the average force of friction in this case?*print*Print*list*Cite

The mass of Billy's automobile is 950 kg. The force provided by the engine is 10.74 kN.

There are two significant forces acting on the automobile which determine its rate of acceleration. One is the force provided by the engine in the forward direction and another is an opposing force in the backward direction due to friction.

The acceleration of the automobile is (change in velocity)/time

=> (27.8 - 0)/8.52

=> 3.2629 m/s^2

If the force due to friction is F, the net force in the forward direction is the product of mass and acceleration

=> 950*3.2629 = 3099.76 N or 3.099 kN

The force provided by the engine is 10.74 kN. So the frictional force is F = 10.74 - 3.099 = 7.641 kN

**The average force of friction is 7.641 kN.**