A 2.98 × 103 kg car requires 6.0 kJ of work tomove from rest to some final speed. Duringthis time, the car moves 29.8 m.Neglecting friction, find...

I understand that the weight of the car specified is intend to be 2.98 x 10^3 kg rather than 2.98 x 10^3 kg, and the problem is solved accordingly.

Given:

Weight of car = m = 2.98 x 10^3 kg

Work done on car = 6.0 kJ = 6000 J

Distance covered by car = s = 29.8 m

The work done on the car has been converted into kinetic energy Which is given by the formula

Kinetic energy of car = 6000 J = (1/2)*m*v^2 =

where v = final speed of car.

Therefore final speed of car = v= [6000*(2/m)]^1/2

= [(2*6000)/(2.98 x 10^3)]^1/2 = 2.0067 m/s

Acceleration of the car is given by the formula:

Acceleration = a =  (v^2)/2*s = (2.0067^2)/(2*29.8) = 0.0676 m/s

And force exerted on car = m*a = 2.98 x 10^3)*0.0676 = 201.3422 N

Answer:

Final speed of car = 2.0067 m/s

Horizontal force exerted on car = 201.3422 N