An engine is attached to a wagon through a shock absorber of length 1.5 m . The system with a total mass of $$50,000 \mathrm{~kg}$$ is moving with a speed of $$36 \mathrm{~kmh}^{-1}$$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m . If $$90 \%$$ of energy of the wagon is lost due to friction, calculate the spring constant.
$$\begin{aligned} \text { Given, mass of the system } (m) & =50,000 \mathrm{~kg} \\ \text { Speed of the system }(v) & =36 \mathrm{~km} / \mathrm{h} \\ & =\frac{36 \times 1000}{60 \times 60}=10 \mathrm{~m} / \mathrm{s} \end{aligned}$$
$$\begin{aligned} \text { Compression of the spring }(x) & =1.0 \mathrm{~m} \\ \qquad \mathrm{KE} \text { of the system } & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} \times 50000 \times(10)^2 \\ & =25000 \times 100 \mathrm{~J}=2.5 \times 10^6 \mathrm{~J} \end{aligned}$$
Since, $$90 \%$$ of KE of the system is lost due to friction, therefore, energy transferred to shock absorber, is given by
$$\begin{aligned} \Delta E & =\frac{1}{2} k x^2=10 \% \text { of total KE of the system } \\ & =\frac{10}{100} \times 2.5 \times 10^6 \mathrm{~J} \text { or } k=\frac{2 \times 2.5 \times 10^6}{10 \times(1)^2} \\ & =5.0 \times 10^5 \mathrm{~N} / \mathrm{m} \end{aligned}$$
An adult weighting 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $$10 \%$$ of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.
$$\begin{aligned} \text { Given, weight of the adult }(w) & =m g=600 \mathrm{~N} \\ \text { Height of each step } & =h=0.25 \mathrm{~m} \\ \text { Length of each step } & =1 \mathrm{~m} \end{aligned}$$
$$\begin{aligned} & \text { Total distance travelled }=6 \mathrm{~km}=6000 \mathrm{~m} \\ & \therefore \quad \text { Total number of steps }=\frac{6000}{1}=6000 \end{aligned}$$
$$\begin{aligned} \text { Total energy utilised in jogging } & =n \times m g h \\ & =6000 \times 600 \times 0.25 \mathrm{~J}=9 \times 10^5 \mathrm{~J} \end{aligned}$$
Since, $$10 \%$$ of intake energy is utilised in jogging.
$$\therefore$$ Total intake energy $$=10 \times 9 \times 10^5 \mathrm{~J}=9 \times 10^6 \mathrm{~J}$$.
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.