# A 2.09 × 10^3 kg car requires 5.7 kJ of work to move from rest to some final speed. During this time, the car moves 23.7 m.Neglecting friction, find the final speed and net horizontal force...

A 2.09 × 10^3 kg car requires 5.7 kJ of work to move from rest to some final speed. During this time, the car moves 23.7 m.

Neglecting friction, find the final speed and net horizontal force exerted on the car.

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### 1 Answer

Work is dot product of force and displacement. A 2.09*10^3 kg car requires 5.7 kJ of work to move from rest to a final speed V. The displacement of the car during this is 23.7 m in the horizontal direction.

As the car starts from rest the initial velocity is u = 0. Use the formula v^2 - u^2 = 2*a*s

=> V^2 = 2*a*23.7

The work done : W = F*d

=> 5.7*10^3 = F*23.7

=> 5.7*10^3 = 2.09*10^3* a* 23.7

=> a = 5.7*10^3/(2.09*10^3*23.7)

=> a = 0.115 m/s^2

v^2 = 2*23.7*0.115 = 5.45 m/s

v = 2.33 m/s

The force is 5.7/23.7 = 0.24 N

The final velocity is 2.33 m/s