On complete combustion a litre of petrol gives off heat equivalent to $$3 \times 10^7 \mathrm{~J}$$. In a test drive, a car weighing 1200 kg including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.
Energy is given by the petrol in the form of heat of combustion.
Thus, by question,
$$\begin{aligned} \text { Energy given by } 1 \text { litre of petrol } & =3 \times 10^7 \mathrm{~J} \\ \quad \text { Efficiency of the car engine } & =0.5 \\ \therefore \quad \text { Energy used by the car } & =0.5 \times 3 \times 10^7 \mathrm{~J} \\ E & =1.5 \times 10^7 \mathrm{~J} \end{aligned}$$
Total distance travelled $$(\mathrm{s})=15 \mathrm{~km}=15 \times 10^3 \mathrm{~m}$$ If $$f$$ is the force of friction then,
$$\begin{aligned} E & =f \times s \quad(\because \text { Energy is utilised in working against friction) } \\ 1.5 \times 10^7 & =f \times 15 \times 10^3 \\ \Rightarrow \quad f & =\frac{1.5 \times 10^7}{15 \times 10^3}=10^3 \mathrm{~N} \\ f & =1000 \mathrm{~N} \end{aligned}$$
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of $$30^{\circ}$$ by a force of 10 N parallel to the inclined surface (figure). The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
(a) work done against gravity
(b) work done against force of friction
(c) increases in potential energy
(d) increase in kinetic energy
(e) work done by applied force
Consider the adjacent diagram the block is pushed up by applying a force $$F$$.
Normal reaction $$(N)$$ and frictional force $$(f)$$ is shown.
Given, mass $$=m=1 \mathrm{~kg}, \theta=30^{\circ}$$
$$F=10 \mathrm{~N}, \mu=0.1$$ and $$\mathrm{s}=$$ distance moved by the block along the inclined plane $$=10 \mathrm{~m}$$
$$\begin{aligned} &\text { (a) } \quad \text { Work done against gravity }=\text { Increase in PE of the block }\\ &\begin{aligned} & =m g \times \text { Vertical distance travelled } \\ & =m g \times s(\sin \theta)=(m g s) \sin \theta \end{aligned}\\ &=1 \times 10 \times 10 \times \sin 30^{\circ}=50 \mathrm{~J} \quad\left(\because g \leq 10 \mathrm{~m} / \mathrm{s}^2\right) \end{aligned}$$
$$\begin{aligned} &\text { (b) Work done against friction }\\ &\begin{aligned} w f & =f \times s=\mu N \times s=\mu m g \cos \theta \times s \\ & =0.1 \times 1 \times 10 \times \cos 30^{\circ} \times 10 \\ & =10 \times 0.866=8.66 \mathrm{~J} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text {(c) Increase in PE }=m g h & =m g(\mathrm{~s} \sin \theta) \\ & =1 \times 10 \times 10 \times \sin 30^{\circ} \\ & =100 \times \frac{1}{2}=50 \mathrm{~J} \end{aligned}$$
$$\begin{aligned} &\text { (d) By work-energy theorem, we know that work done by all the forces = change in } \mathrm{KE}\\ &\begin{aligned} (W) & =\Delta K \\ \Delta k & =W_g+W_f+W_f \\ \Rightarrow \quad & =-m g h-f s+F S \\ & =-50-8.66+10 \times 10 \\ & =50-8.66=41.34 \mathrm{~J} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { (e) Work done by applied force, } F=F S\\ &=(10)(10)=100 \mathrm{~J} \end{aligned}$$
A curved surface is shown in figure. The portion $$B C D$$ is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from $$A$$ which is at a slightly greater height than $$C$$.
With the surface $$A B$$, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(a) For which balls is total mechanical energy conserved?
(b) Which ball (s) can reach D?
(c) For balls which do not reach $$D$$, which of the balls can reach back $$A$$ ?
(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved.
Ball 3 is having negligible friction hence, there is no loss of energy.
(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.
(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in "wrong" sense when it reaches B. It cannot roll back to A, because of kinetic friction.
$$A$$ rocket accelerates straight up by ejecting gas downwards. In a small time interval $$\Delta t$$, it ejects a gas of mass $$\Delta m$$ at a relative speed $$u$$. Calculate KE of the entire system at $$t+\Delta t$$ and $$t$$ and show that the device that ejects gas does work $$=(1 / 2) \Delta m u^2$$ in this time interval (negative gravity).
Let $$M$$ be the mass of rocket at any time $$t$$ and $$v_1$$ the velocity of rocket at the same time $$t$$.
Let $$\Delta m=$$ mass of gas ejected in time interval $$\Delta t$$.
Relative speed of gas ejected $$=u$$.
Consider at time $$t+\Delta t$$
$$\begin{aligned} (\mathrm{KE})_t+\Delta t & =\mathrm{KE} \text { of rocket }+\mathrm{KE} \text { of gas } \\ & =\frac{1}{2}(M-\Delta m)(v+\Delta v)^2+\frac{1}{2} \Delta m(v-u)^2 \\ & =\frac{1}{2} M v^2+M v \Delta v-\Delta m v u+\frac{1}{2} \Delta m u^2 \\ (\mathrm{KE})_t & =\mathrm{KE} \text { of the rocket at time } t=\frac{1}{2} M v^2 \\ \Delta \mathrm{K} & =(\mathrm{KE})_t+\Delta t-(\mathrm{KE})_t \\ & =(M \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^2 \end{aligned}$$
Since, action-reaction forces are equal.
$$\begin{aligned} &\begin{aligned} \text { Hence, } \quad M \frac{d v}{d t} & =\frac{d m}{d t}|u| \\ \Rightarrow \quad M \Delta v & =\Delta m u \\ \Delta K & =\frac{1}{2} \Delta m u^2 \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Now, by work-energy theorem, }\\ &\begin{aligned} & \Delta K=\Delta W \\ \Rightarrow \quad & \Delta W=\frac{1}{2} \Delta m u^2 \end{aligned} \end{aligned}$$
Two identical steel cubes (masses 50 g, side 1 cm ) collide head-on face to face with a space of $$10 \mathrm{~cm} / \mathrm{s}$$ each. Find the maximum compression of each. Young's modulus for steel $$=Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$$.
$$\begin{aligned} &\begin{aligned} \text{Let} \quad m & =50 \mathrm{~g}=50 \times 10^{-3} \mathrm{~kg} \\ \text { Side } & =L=1 \mathrm{~cm}=0.01 \mathrm{~m} \\ \text { Speed } & =v=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s} \\ \text { Young's modulus } & =Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \end{aligned}\\ &\text { Maximum compression } \Delta L=\text { ? } \end{aligned}$$
Maximum compression $$\Delta L=$$ ?
In this case, all KE will be converted to PE By Hooke's law,
$$\frac{F}{A}=Y \frac{\Delta L}{L}$$
where $$A$$ is the surface area and $$L$$ is length of the side of the cube. If $$k$$ is spring or compression constant, then
$$\begin{aligned} \text { force } F & =k \Delta L \\ \therefore \quad k & =Y \frac{A}{L}=Y L \end{aligned}$$
$$\begin{aligned} & \text { Initial } K E=2 \times \frac{1}{2} m v^2=5 \times 10^{-4} \mathrm{~J} \\ & \text { Final } P E=2 \times \frac{1}{2} k(\Delta L)^2 \end{aligned}$$
$$\therefore \quad \Delta L=\sqrt{\frac{K E}{k}}=\sqrt{\frac{K E}{Y L}}=\sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.1}}=1.58 \times 10^{-7} \mathrm{~m} \quad[\because P E=K E]$$