Question

What is the work required to increase the velocity of car from 36 km/h to 108 km/h, if the mass of the
car is 1200 kg?​

Answer 1

Answer:

W1=mu²/2= (1200×100) /2 = 60000N

W2=mv²/2 = (1200×30²)/2= (1200×900)/2= 540000N

so, W2-W1=work done and your answer

so, m/2(v²-u²)= 540000-60000= 480000N

so the answer=480000N

plz mark me as brainlest and follow me and I would do the same to you

Answer 2

Answer:

The work required to increase the velocity of a car of mass 1200 kg from 36 km/h to 108 km/h will be $480000$  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)

According to the work-energy theorem,

The work done will be equivalent to the change in kinetic energy of the car.

that is,

Work done =  $\Delta K.E$

From the question,

Mass of the car (m) = $1200$ kg

Initial speed of the car  (u)      =  $36$ km/h

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

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

$u = 36 \times \frac{5}{18}$ $= 10\ m/s$

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

Substitutes the values into equation(2).

$\Delta KE = \frac{1}{2}\times1200\times( 30^{2} - 10^{2} )$

        $=480000$ J

So, the amount of work required $=480000$ J