Question

A car weighing 1200 kg is uniformly accelerated from rest and covers a distance of 40 m in 5 s. calculate the work done by the engine of car during this time. What is the final kinetic energy of car

Answer 1

s=ut+1/2at2
40=1/2a(5)^2
a=80/25
a=3.2m/s^2
f=ma
  =1200*3.2
  f=3840N
work=fs
         =3840*40
      w=153600J

Answer 2

Given :

mass of the car (m) = 1200 Kg

Distance covered (s) = 40 m

Time taken (T) = 5 s

To Find :

(i) Work done by the car

(ii) Final Kinetic Energy of the car.

Solution :

(i) As the car is uniformly accelerated, we can use the kinematic equation,

  s = ut + $\frac{1}{2}$at²

Where, s = distance covered; u = initial velocity; t = time taken; a = acceleration

∴    s    = ut  +  $\frac{1}{2}$at²

⇒  40  = 0×5 + $\frac{1}{2}$×a×5²

⇒  40  =    $\frac{1}{2}$×a×25

⇒  80  =  25a

⇒    a   =  $\frac{80}{25}$

∴     a   =  3.2 m/sec×

We know,

     Work Done (W) = Force (F) × Displacement (s)

⇒  Work Done (W) = ma × s

⇒  Work Done (W) = ma × s

⇒  Work Done (W) = 1200 × 3.2 × 40

∴   Work Done (W) = 153600 J or 153.6 KJ

(ii) Using Kinematic equation we have

        v = u + at             (where v = final velocity)

⇒     v = 0 + 3.2 × 5

∴      v = 16 m/sec

Final Kinetic Energy = $\frac{1}{2}$mv²

                                 = $\frac{1}{2}$ × 1200 × 16²

                                 = 153600 J

Therefore, The work done is 153600 J and Kinetic Energy is also 153600 J

Note : We can find the final kinetic energy using Work-Energy theorem also.