Question

A car of mass 500 kg increases its speed from 72 km/h to 108 km/h. Find the increase in its kinetic energy?​

Answer 1

Answer:

Yes! Here is your answer!

Explanation:

Answer 2

Answer:

A car of mass 500 kg increases its speed from 72 km/h to 108 km/h. The increase in its kinetic energy will be 125000 J  

Explanation:

We have the equation for kinetic energy (K.E) as,

$K.E = \frac{1}{2} mv^{2}$ ...(1)

where m is the mass of the moving object and v is the velocity of the object.

Using equation (1),

Initial K.E   $= \frac{1}{2} mu^{2}$

Final K.E     $= \frac{1}{2} mv^{2}$

The change or increase in kinetic energy ($\Delta K.E$) can be written as,

 $\Delta K.E =\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}$

            $= \frac{1}{2} m(v^{2} - u^{2})$  ...(2)

From the question,

Mass of the car (m) = $50$ kg

Initial speed of the car  (u)      =  $72$ km/h

Final speed of the car  (v)       =  $108$ km/h

Convert km/h into m/s by multiplying the velocity with $\frac{5}{18}$

$u = 72 \times \frac{5}{18}$ $= 20\ m/s$

$v = 108 \times \frac{5}{18}$$= 30 m/s$

Substitutes the values into equation(2).

$KE = \frac{1}{2}\times500\times( 30^{2} - 20^{2} )$

        $= 125,000$ J