# 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 a) the final speed.Answer in units of m/s. b)...

A 2.98 × 103 kg car requires 6.0 kJ of work to

move from rest to some final speed. During

this time, the car moves 29.8 m.

Neglecting friction, find

a) the final speed.

Answer in units of m/s.

b) the net horizontal force exerted on the car.

Answer in units of N.

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### 2 Answers

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

The energy required by the car to attain a speed (velocity) v is (1/2)mv^2, where m is the mass of the car, the mass of the car Given, m=2.98*103 kg and the energy 6kJ =6000J.

Therefore, (1/2)(2.98*103)*v^2=6000. So,

v=sqrt[6000*2/(2.98*103)]= 6.252646537m/s

Therefore, the acceleration , a of the car is given by the equation of the motion:

v^2-u^2=2as, , where s is the displacement of the car . u and v are its initial and final velocities. But u= 0 at the starting and v=6.252646537m/s, as calculated, s=29.8m given.

Therefore, a=(V^2-u^2)/(2s)=(6.252646537m/s)^2-(0 m/s)^2]/29.8m = 1.311932507m/s^2.

b)

The horizontal force exerted on the car = its mass*acceleration =2.98*103kg* 1.311932507m/s^2.402.6846 N