Question

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 bottom of the driveway is 3.8 m/s. What is the length of the driveway?

Answer 1

Answer:

The Length of the driveway is   4.98 m

Explanation:

we have to determine the length of the driveway  

so we use following equations

W=ΔK.E    where W is work and  ΔK.E   change in the kinetic energy

also

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

also

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

Given that

$F_{f}$= 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

$W = (1/2)(2100)(3.8)^2$ =  15162  J  

Since the x component of the gravity is

$Fx_$ = mg sinФ

so

$Fx_{}$ = (2100)(9.8)sin(20.0°  ) we get

$Fx_{}$ = 7038.77 N

And the Net force is

$F_{net}$ = $Fx_ {}$ - $F_{f}$

$F_{net}$ = 7038.77 - 4000 = 3038.77 N

And the length of the driveway = W / ($F_{net}$ ) = 15162/3038.77 = 4.98 m

So this the length of the driveway.