Which one needs more work to be done?

The amount of work is equal to the change in kinetic energy, $K \propto v^2$ . If the kinetic energy at 50 km/h is $K_0$ , then the kinetic energy at 100 km/h is $4K_0$ .

Therefore the amount of work in the first part is $K_0 - 0 = K_0$ , and in the second part $4K_0 - K_0 = 3K_0$ , which is three times $\boxed{\text{more}}$ .

The work done is $W = Fd = mad$ , where $m$ is the mass of the car, $a$ the (assumed constant) acceleration and $d$ the distance over which the car accelerates.

Now $v_{f}^{2} = v_{i}^{2} + 2ad \Longrightarrow d = \dfrac{v_{f}^{2} - v_{i}^{2}}{2a} \Longrightarrow W = mad = ma \dfrac{v_{f}^{2} - v_{i}^{2}}{2a} = \dfrac{1}{2}m(v_{f}^{2} - v_{i}^{2})$ ,

so since $100^{2} - 50^{2} = 3 \times (50^{2} - 0^{2})$ , the work done going $\boxed{\text{from 50 km/hr to 100 km/hr}}$ will be more, (in fact 3 times more), than going from rest to 50 km/hr.

Note that, in fact, we do not need to make any assumptions about the constancy of the acceleration. Any work done by the engine will go into the kinetic energy $E_{k}$ of the car, (assuming that the road is level). Since $E_{k} = \dfrac{1}{2} mv^{2}$ and $W = \Delta E_{k}$ , and since again $100^{2} - 50^{2} = 3(50^{2} - 0^{2})$ , the work done going from 50 km/hr to 100 km/hr is greater.

Using the Kinetic energy theoreme $\frac { 1 }{ 2 } m{ { v }_{ 1 } }^{ 2 }-\frac { 1 }{ 2 } m{ { v }_{ 2 } }^{ 2 }=W\left( \overset { \rightarrow }{ F } \right)$ we get $\frac { W\left( \overset { \rightarrow }{ { F }_{ 2 } } \right) }{ W\left( \overset { \rightarrow }{ { F }_{ 1 } } \right) } =\frac { { 100 }^{ 2 }-{ 50 }^{ 2 } }{ { 50 }^{ 2 }-{ 0 }^{ 2 } } =3$ . So the work done from 50km/h to 100km/h is three times the work done from 0km/h to 50km/h.

We know that, Work $=$ change in kinetic energy $=$ $K.E(final) - K.E(initial)$ . [ Work energy theorem]

And, kinetic energy $=$ $\frac{1}{2}$ . $m$ . $v^{2}$ . Where, $m = mass$ and $v = velocity$ .

$\therefore$ In the first case, we have $K.E(initial) = \frac{1}{2}.m.0^{2}$ $=$ $0$ .

And, $K.E(final)$ $=$ $\frac{1}{2}$ . $m$ . $50^{2}$

Which is $\frac{2500}{2}$ . $m$ $=$ $1250$ $m$ .

So, The work done in the first case, $K.E(final) - K.E(initial)$ $=$ $1250$ . $m-$ $0$ .

As the mass is constant in both the cases, we neglect it, so the work done $=$ $\boxed{1250}$ .

Similarly, in the second case, we have $K.E(final) - K.E(initial)$ $=$ $\frac{1}{2}$ .( $100^{2} - 50^{2}$ )

Which is $\frac{7500}{2}$ $=$ $\boxed{3750}$ .

So, work done in the second case is greater!

$3750 > 1250$

$\Rightarrow$ The work done in the second case is 3 times greater than the work done in the first case!

Consider the car accelerate with a constant acceleration of $a$ from an initial velocity of $u$ to final velocity of $v$ , then the displacement $s$ of the car is given by $2as = v^2 - u^2$ , neglecting air drag.

Since work done $W=Fs$ , where $F$ is the force exerted by the car engine, if the mass of the car be $m$ , then $W=Fs = mas = \dfrac {m(v^2-u^2)}2$ $\implies W \propto v^2-u^2 = k(v^2-u^2)$ , where $k$ is a constant. Therefore, $W_1 = (50^2-0^2)k = 2500k$ and $W_2 = (100^2-50^2)k = 7500k$ . Therefore, $W_2$ or work for car accelerating from 50 km/hr to 100 km/hr is more.

We know that: $W=K_{2} - K_{1} \Leftrightarrow W=\frac {1}{2}mu^2_2-\frac {1}{2}mu^2_1$ where $u_2,u_1$ the final and starting velocities of an object respectively. If we input $50$ and $0$ for $u_2$ and $u_1$ respectively we get $W_1=-2500\frac {1}{2}m$ . Whereas inputing $100$ and $50$ for $u_2$ and $u_1$ respectively gives a result of $W_2=-7500\frac {1}{2}m$ . Therefore we conclude that $W_2=3W_1$ so the Work is greater $\boxed {\text{From 50 km/hr to 100 km/hr}}$

Work done = change in kinetic energy ( from work energy theorem) m/2 (50^2 - 0) < m/2 (100^2 - 50^2) so in second case more work is done

From rest the car needs a force to put it in motion, therefore when it is in motion it needs no much force again since it is already in its kinetic state, so therefore the engine performs with less force making it to go faster easy with less effort.