A NASCAR of mass 1400kg has broken down on pit lane. The coefficient of static friction between the tires and the road is 0.55. How hard would the crew have to push to accelerate the car at 3.60m/s²?

1 Answer
A08
Mar 16, 2018

Force of friction is given as

#vec(F_f)=muvecN#
where #mu# is coefficient of static friction and #vecN=mvecg# is normal reaction. Here #vecg=9.81\ ms^-2# is acceleration due to gravity

Force required to produce given acceleration in the car is found using Newton's Second Law of motion

#vecF=mveca#

Total force required to be applied by the crew

#vecF_"Total"=mveca+muvecN#
#=>vecF_"Total"=mveca+mumvecg#
#=>vecF_"Total"=m(veca+muvecg)#

Taking direction of car's acceleration and of force of friction to be same, inserting given vales we get

#F_"Total"=1400(3.60+0.55xx9.81)#
#F_"Total"=12593.7\ N#

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