Correct answer is (3)

The kinetic energy of a moving object of mass $m$ and velocity $v$ is given by the formula:

$K=\frac{1}{2}m{v}^2$

The work done in accelerating an object from rest to velocity $v$ is equal to its change in kinetic energy. Therefore, the energy spent in accelerating the car from rest to $u\mathrm{m}/\mathrm{s}$ is:

$E=\frac{1}{2}m{u}^2$

The energy required to accelerate the car from $u\mathrm{m}/\mathrm{s}$ to $2u\mathrm{m}/\mathrm{s}$ is:

$\begin{array}{rl}nE& =\frac{1}{2}m(2u{)}^2-\frac{1}{2}m{u}^2\\ \\ & =2m{u}^2-\frac{1}{2}m{u}^2\\ \\ & =\frac{3}{2}m{u}^2\\ \\ & =3E\end{array}$

$\therefore$ n = 3

Correct answer is (784)

To maintain a constant speed, the bus has to overcome the frictional force acting on it. The frictional force is given by:

$F_{friction}=\mu F_N$

Where $\mu$ is the coefficient of friction and $F_N$ is the normal force acting on the bus. Since the bus is on a flat road, the normal force is equal to the gravitational force:

$F_N=mg$

Where $m$ is the mass of the bus and $g$ is the acceleration due to gravity (approximately $9.8\mathrm{m}/{\mathrm{s}}^2$).

Substituting the values, we get:

$F_{friction}=0.04\times 500\times 9.8$

$F_{friction}=196\mathrm{N}$

To maintain a constant speed, the engine must exert a force equal in magnitude to the frictional force. The work done by the engine to overcome the frictional force is given by:

$W=F_{friction}\times d$

Where $d$ is the distance traveled. First, convert the distance from km to m:

$d=4\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}=4000\mathrm{m}$

Now, calculate the work done:

$W=196\mathrm{N}\times 4000\mathrm{m}$ $W=784000\mathrm{J}$

Convert the work done from joules to kilojoules:

$W=\frac{784000\mathrm{J}}{1000\mathrm{J}/\mathrm{kJ}}=784\mathrm{kJ}$

The work done by the engine of the bus to maintain a speed of 80 km/h for a 4 km distance is $784.8\mathrm{kJ}$.

Correct answer is (245)

From work energy theorem

$\begin{array}{rl}& W_{\text{gravity }}+W_{\text{friction }}=\mathrm{\Delta }(KE)=K{E}_f-K{E}_i\\ \\ & W_{\text{gravity }}=mgh=1\times 10\times 0.3=3\mathrm{J}\\ \\ & W_{\text{friction }}=0-\frac{1}{2}\times (22{)}^2-3\\ \\ & =-(242+3)=-245\mathrm{J}\end{array}$

Correct answer is (30)

The increase in kinetic energy can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done by a force is given by the equation:

$W=F\cdot d$

where ( F ) is the force and ( d ) is the distance over which the force is applied.

However, we don't have the distance in this problem. But we do know that the force is applied for a time of 5 seconds, and that the initial momentum of the body is 10 kg m/s. We can use these facts to find the work done.

First, we can use the equation for force, ( F = ma ), to find the acceleration of the body:

$a=\frac{F}{m}=\frac{2\text{N}}{5\text{kg}}=0.4{\text{m/s}}^2$

Then, we can use the equation for distance in uniformly accelerated motion, ( $d=v_{i}t+\frac{1}{2}a{t}^2$), where ( $v_i$ ) is the initial velocity of the body. We can find ( $v_i$) from the initial momentum and the mass of the body:

$v_i=\frac{p}{m}=\frac{10\text{kg m/s}}{5\text{kg}}=2\text{m/s}$

Substituting ( $v_i$ ), ( a ), and ( t ) into the equation for ( d ) gives:

$d=2\text{m/s}\cdot 5\text{s}+\frac{1}{2}\cdot 0.4{\text{m/s}}^2\cdot (5\text{s}{)}^2=10\text{m}+5\text{m}=15\text{m}$

Finally, we can substitute ( F ) and ( d ) into the equation for work to find the increase in kinetic energy:

$\mathrm{\Delta }KE=W=F\cdot d=2\text{N}\cdot 15\text{m}=30\text{J}$

So, the increase in the kinetic energy of the body is 30 J.

Correct answer is (375)

The velocity of the body just before hitting the ground, due to gravitational acceleration, is given by $v_1=\sqrt{2g{h}_1}$, and the velocity just after hitting the ground, when it rebounds to a height $h_2$, is given by $v_2=\sqrt{2g{h}_2}$.

According to the problem, the ratio $\frac{v_1}{v2}=4$. Therefore, we can write $\frac{\sqrt{2g{h}_1}}{\sqrt{2g{h}_2}}=4$ or equivalently $\frac{h_1}{h_2}=4^2=16$.

The loss in kinetic energy due to the collision with the ground is given by the difference between the initial kinetic energy $K_1=\frac{1}{2}m{v}_1^2$ and the final kinetic energy $K_2=\frac{1}{2}m{v}_2^2$, where m is the mass of the body.

Substituting $v_1=\sqrt{2g{h}_1}$ and $v_2=\sqrt{2g{h}_2}$ into these expressions, we get $K_1=mg{h}_1$ and $K_2=mg{h}_2$.

The loss in kinetic energy is then $\mathrm{\Delta }K=K_1-K_2=mg{h}_1-mg{h}_2$.

The percentage loss in kinetic energy is given by

$\frac{\mathrm{\Delta }K}{K_1}\times 100=\frac{mg{h}_1-mg{h}_2}{mg{h}_1}\times 100=\frac{h_1-h_2}{h_1}\times 100$.

Since $h_1/h_2=16$, we can write $h_2=h_1/16$, so the percentage loss in kinetic energy is

$\frac{h_1-h_1/16}{h_1}\times 100=100(1-\frac{1}{16})=100\times \frac{15}{16}=\frac{375}{4}$.

So, the value of $x$ is 375.