Question

calculate the work required to be done to stop a car of l500 moving at a velocity of 72 km/h

Answer 1

Work done to stop a car = kinetic energy of the car .

Formula

Kinetic energy = 1/2 × m v²

Given :

Mass ( m ) = 1500 kg

velocity ( v ) = 72 km/hr

K = 1/2 m v²

  = 1/2 × ( 1500 kg ) × ( velocity )²

Hold on !

72 km/hr should be converted to m/s

72 km = 72000 m

1 h = 3600 s

72 km/hr = 72,000 / 3600 m/s

= > 20 m/s

Now : K = 1/2 × 1500 × ( 20 )² J

= > K = 750 × 400 J

= > K = 3,00,000 J

ANSWER :

The work done is 3.0 × 10⁵

Answer 2

Work done = Change in kinetic energy.

So, in this case, Initial velocity (u) = 72km/h= 20m/s
and, final velocity (v)= 0m/s
So, change in kinetic energy will be
$\frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} \\ = \frac{1}{2} m(0 - 400) \\ = \frac{1}{2} \times 1500 \times - 400 \\ = - 300000 \: joules$
Here you can ignore the minus sign as it signifies that the force is not a repulsive force