Question

the speed of a vehicle of mass 500kg increases its velocity from 56km/hr to 72km/hr.what is the increase in kinetic energy.

Answer 1

The task:

the speed of a vehicle of mass 500kg increases its velocity from 56km/hr to 72km/hr.what is the increase in kinetic energy.

Solution:

56 km/hr=16 m/sec

72 km/hr=20 m/sec

A kinetic energy of a vehicle before increasing its velocity is

$$E _ {1} = \frac {1}{2} m v _ {1} ^ {2}$$

And after increasing of the velocity, kinetic energy of the vehicle is

$$E _ {2} = \frac {1}{2} m v _ {2} ^ {2}$$

So the increase in kinetic energy is

$$\Delta E = E _ {2} - E _ {1}$$

Or

$$\Delta E = \frac {m}{2} (v _ {2} ^ {2} - v _ {1} ^ {2})$$ $$\Delta E = \frac {500\ \text{kg}}{2} (20^2 - 16^2) = 36000\ \text{J}$$

The answer: $\Delta E = 36\ \text{kJ}$