A car accelerates down a hill that is #45.0^@# to the horizontal. If the car has a #350kW# engine and takes #2.4s# to reach the bottom, what is the force of friction resisting its motion?

A car of #750kg# accelerates down a hill. At the top of the hill, its speed is #75(km)/h#, while at the bottom, its speed is #185(km)/h#. The hill is #40.0m# long and makes an angle of #45.0^@# with the horizontal. If the car has a #350kW# engine and takes #2.4s# to reach the bottom, what is the force of friction resisting its motion?

1 Answer
Jun 24, 2016

Answer:

#5508.575N

Explanation:

self drawn

Given

  • # m->"Mass of the car"= 750kg#
  • #v_i-> "Initial velocity of the car at top of the hill"= 75"km"/h#
    #=(75xx1000)/3600m/s=375/18 m/s#
  • #v_f-> "Final velocity of the car at hill bottom "= 185"km"/h#
    #=(185xx1000)/3600m/s=925/18 m/s#
  • #d -> " Length of the hill path"=40 m#
  • #a-> "Angle of inclination of path "=45^@#
  • #P->"The power of car engine"=350kW=350xx10^3W#
  • #t -> "The time descent from top to bottom"= 2.4 s#
  • #F->"Force of friction (in N)"=?#

We are to apply the law of conservation of energy to solve the problem in the easiest way.

Here the net gain in KE of car will be equal to the sum of the enegy supplied by the engine and work done by the gravitational pull minus the loss of energy due to friction( i.e. work done against force of friction)

Let us calculate each separately

  • #"Total gain in KE" ,DeltaE_k=1/2m(v_f^2-v_i^2)#

#=1/2xx750xx((925/18)^2-(375/18)^2)J#

#=375xx(1300xx550)/(18xx18)=827546.3J#

  • #E_"engine"->"Energy supplied by engine"=Pxxt#

#E_"engine"=350000xx2.4J=840000J#

  • #W_g->"work done by gravitational pull"=mgsinaxxd#

#=750xx9.8xxsin45xx40J=207889.3J#

  • #W_"friction"=Fxxd=Fxx40J#

So by the law of conservation of energy

#DeltaE_k=E_"engine"+W_g-W_"friction"#

#=>W_"friction"=E_"engine"+ W_g -DeltaE_k#

#=>Fxx40=840000+207889.3-827546.3=220343#

#:.F=220343/40=5508.575N#