Question

A 1500-kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s over a dry concrete road. Assume that kinetic friction remains constant at 400 N during this time. –

(a) Find the net work done on the car and the work done by the engine. –

(b) How many revolutions does each tire make over the 12.0 s interval? What is the angular speed of the wheels when the car has traveled half the total distance? Assume that the tires have radii of 0.330 m. Note that 1 rev = 2π rad. –

(c) Suppose that the engine shuts off after 12 s. If the car comes to rest after traveling a distance of 50 m over a wet concrete surface, determine the coefficient of friction between the wet concrete surface and the car.

Answer 1

a) Net workdone on the car = change in kinetic energy

= 0.5*m*(v^2 - u^2)

= 0.5*1500*(18^2 - 0^2)

= 243000 J

acceleration of the car, a = (v - u)/t

= (18 - 0)/12

= 1.5 m/s^2

distance travelled during 12s,

d = u*t + 0.5*a*t^2

= 0*12 + 0.5*1.5*12^2

= 432 m

Workdone by the engine = 243000 + 400*432

= 415800 J

b) no of turns rotated, N = d/(2*pi*r)

= 432/(2**pi*0.33)

= 208

c) acceleration, a = (0^2 - 18^2)/(2*50)

= -3.24 m/s^2

we know, a = -g*mue_k

==> mue_k = -a/g

= -(-3.24)/9.8

= 0.331