Question

1. A 2.1 × 103 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0° with the horizontal. An average friction force of 4.0 × 103 N impedes the car’s motion so that the car’s speed at the bot- tom of the driveway is 3.8 m/s. What is the length of the driveway?

Answer 1

$K.E=\frac{mv^2}{2}$

also

W = F.d Where F is Force and d is distance

Given that

F= 4000 N this frictional force

m = 2100 Kg

θ= 20.0°

V=3.8 m/s V is the car’s speed at the bottom of the driveway

W=Δ K.E

 $W = 0.5×2100×3.8^2=15162 J$

Since the x component of the gravity is

$Fx = mg sinФ$

so

$Fx = 2100×9.8sin(20.0° ) =7038.77 N$

And the Net force is

 $F_{net}=F_x-F_f$

$F_{net}= 7038.77 – 4000 = 3038.77 N$

And the length of the driveway =$\frac{W}{F_{net}} =\frac{15162}{3038.77} = 4.98 m$

So the length of the driveway=4.98M