Work, Energy and Power

Q: A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.(a) How much work does she do against the gravitational force?(b) Fat supplies 3.8 \$\times\$ \$10^7\$ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

(a) Total work done = \$mgh\times \text {no. of times} =10\times 9.8 \times 0.5 \times 1000=49000J\$(b) Energy equivalent of 1 kg of fat \$= 3.8 \times 10^7 J\$Efficiency rate = 20%Mechanical energy supplied by the persons body:= \$(20/100) \times 3.8\times 10^7 J\$= \$(1/5)\times 3.8 \times 10^7\$ JEquivalent mass of fat lost by the dieter:\$ = \dfrac{1}{\dfrac{1}{5} \times 3.8 \times 10^7} \times 49 \times 10^3\$\$=(245 / 3.8) \times 10\$ \$^{-4}= 6.45 \times 10^{-3} kg\$

Q: The blades of a windmill sweep out a circle of area A.(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?(b) What is the kinetic energy of the air?(c) Assume that the windmill converts 25% of the winds energy into electrical energy, and that A = 30 \$m^2\$, v = 36 km/h and the density of air is 1.2 kg \$m^{3}\$. What is the electrical power produced?

(a) Density of air \$= \rho \ \ kg /m^3\$Time taken = \$t\$ secVelocity of air = \$v\$ m/sMass of the air = density \$\times \$ volume \$= \rho A v t\$(b) \$KE = \cfrac{1}{2}mv^2\$ \$ = \cfrac{1}{2}(\rho A vt) v^2\$ \$ = \cfrac{1}{2}(\rho A v^3t) \$(c) Velocity, v= 36 km/hr = 10 m/s\$\rho = 1.2 kg/m^3\$\$A = 30 m^2\$Work done = change in kinetic energy\$ = \dfrac{1}{2} \times 1.2 \times 30 \times 10^3 t \,Nm\$\$ = 18000t \,J\$Power = \$work /t = 18000 W\$Efficiency = 25%Actual power\$ = 18000 \times 25/100 = 4500 W\$

Q: The potential energy function for a particle executing linear simple harmonic motion is given by \$V(x) = kx^2/2\$, where k is the force constant of the oscillator. For \$k = 0.5 N m^{-1}\$, the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches \$x = \pm2 m\$.

Total energy of the particle, \$E = 1 J\$Force constant, \$k = 0.5 N/ m\$Kinetic energy of the particle, \$K = (1/2)mv^2\$According to the conservation law:\$E = V + K1 = (1/2)kx^2 + (1/2)mv^2\$At the moment of turn back, velocity (and hence K) becomes zero.\$1 = (1/2)kx^2\$\$(1/2) 0.5x^2 = 1\$\$x = 2\$Hence, the particle turns back when it reaches \$x = 2 m.\$

Q: A bullet of mass \$0.012\ kg\$ and horizontal speed \$70\ ms^{-1}\$ strikes a block of wood of mass \$0.4\ kg\$ and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Mass of the bullet, m = 0.012 kgInitial speed of the bullet, \$u_b = 70 m/s\$Mass of the wooden block, \$M = 0.4 \ kg\$Initial speed of the wooden block, \$u_B\$ =0Final speed of the system of the bullet and the block = \$v \ \ m/s\$Applying the law of conservation of momentum:\$m u_b + M u_B = (m + M) v \$\$0.012 \times 70 + 0.4 \times 0 = (0.012 + 0.4) v\$\$v = 0.84 / 0.412\$ \$ =2.04\$ m/sFor the system of the bullet and the wooden block:Mass of the system, \$m' = 0.412 \ kg\$Velocity of the system \$=2.04 m/s\$Height up to which the system rises = \$h\$Applying the law of conservation of energy to this system:Potential energy at the highest point \$=\$ Kinetic energy at the lowest point\$m'gh = (1/2)m'v^2\$\$h = \dfrac{v^2 }{2g}\$\$=\dfrac{ (2.04)^2}{2 \times 9.8} \$\$= 0.2123 m \$The wooden block will rise to a height of \$0.2123 m\$.The heat produced = Kinetic energy of the bullet - Kinetic energy of the system\$= (1/2) mu^2 - (1/2) m'v^2\$ \$=(1/2) \times 0.012 \times (70)^2 - (1/2) \times 0.412 \times (2.04)^2\$\$= 29.4 -0.857 = 28.54 J\$

Q: A rain drop of radius \$2 mm\$ falls from a height of \$500 m\$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on touching the ground is \$10 ms^{-1}\$

Given:Radius of the rain drop \$r= 2mm = 2 \times 10^{-3}m\$Density of water, \$\rho = 10^3 kg m^{-3}\$ The mass contained in a rain drop, \$m = \rho V \$\$\Rightarrow \dfrac{4}{3} \times 3.141 \times (2 \times 10^{-3})^3 \times 10^3 kg \$ Gravitational force experienced by the rain drop,\$F= mg\$\$ \Rightarrow \dfrac{4}{3} \times 3.141 \times ((2 \times 10^{-3})^3 \times 10^{3} \times 9.8) N\$The work done by the gravity on the drop is given by:\$W_1 = Fs \$\$= \dfrac{4}{3} \times 3.14 \times (2 \times 10^3)^3 \times 10^3 \times 9.8 \times 250= 0.082 J\$The work done on the drop in the second half of the journey will be same as that in the first half. \$\therefore W_2=0.082\ J\$The total energy of the drop remains conserved during its motion.Total energy at the top:\$E_T= mgh + 0 \$\$= 4/3 \times 3.141 \times (10^{-3})^3 \times 10^3 \times 500 \times 10^{-5}\$\$= 0.164J\$Due to the presence of a resistive force, the drop hits the ground with a velocity of \$10 m/s\$.∴Total energy at the ground:\$E_G= \dfrac{1}{2}mv^2+0 = 1/2 \times 4/3 \times 3.141 \times (2 \times 10^{-3})^3 \times 10^3 \times 9.8 \times 10^2= 1.675 \times 10^{-3} J \$So, the work done by the resistive force = \$E_G- E_T= -0.162 J\$

Q: A pump on the ground floor of a building can pump up water to fill a tank of volume \$30m^3\$ in \$15 min\$. If the tank is \$40 m\$ above the ground, and the efficiency of the pump is \$30\%\$, how much electric power is consumed by the pump ?[Dencity of water \$10^3kgm^{-3}\$]\$(g=9.8ms^{-2})\$

Given:Volume of the tank, \$V = 30 m^3\$Time of operation, \$t = 15 min = 15 \times 60 = 900 s\$Height of the tank, \$h = 40 m\$Efficiency of the pump, \$= 30\$ %Density of water, \$=10^3 kg/m^3\$Mass of water, \$m = 30 \times 10^3kg\$Output power can be obtained as:\$P_0 = \dfrac{Work \ done}{ Time} = \dfrac{mgh}{t}\$ \$= \dfrac{30 \times 10^3 \times 9.8 \times 40 }{ 900} = 13.067 \times 10^3 W\$For input power \$P_i\$, efficiency is given by the relation:\$\dfrac{P_o}{P_i} = 30\$%\$P_i = \dfrac{13.067 \times 100 \times 10^3 }{30}\$\$P_i=43.6 kW\$

Q: A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the(a) work done by the applied force in 10 s.(b) work done by friction in 10 s,(c) work done by the net force on the body in 10 s,(d) change in kinetic energy of the body in 10 s,and interpret your results.

Given:The mass of the body is \$ m = 2 kg\$The force applied on the body \$F = 7 N\$The coefficient of kinetic friction \$= 0.1\$Since the body starts from rest, the initial velocity of body is zero.Time at which the work is to be determined is \$ t = 10 s\$The acceleration produced in the body by the applied force is given by Newtons second law of motion as:\$a' =\dfrac F m = \dfrac72 = 3.5 m/s^2\$Frictional force is given as:\$ f= \mu g = 0.1\times 2\times 9.8 = 1.96 \$The acceleration produced by the frictional force:\$a" = -\dfrac{1.96}{ 2} = -0.98 m/s^2\$Therefore, the total acceleration of the body: \$a' + a"= 3.5 + (-0.98) = 2.52 m/s^2\$The distance traveled by the body is given by the equation of motion:\$s = ut +\dfrac12at^2\$\$= 0 + \dfrac12 \times2.52 \times (10) ^2=126\ m\$(a) Work done by the applied force, \$W_a = F\cdot s = 7 \times 126 = 882\ J\$(b) Work done by the frictional force, \$W_f = F\cdot s = 1.96\times 126 = 247\ J\$(c), (d)From the first equation of motion, final velocity can be calculated as:\$v = u + at\$\$= 0 + 2.52\times 10 = 25.2 m/s\$So, the change in kinetic energy is \$\Delta K= \dfrac12 mv^2 - \dfrac12 mu^2\$\$= \dfrac12 2(v^2 - u^2) = (25.2) ^2 -0^2 = 635\ J \$The distance traveled by the body is given by the equation of motion:\$s = ut +\dfrac12 at^2= 0 + \dfrac12 \times 2.52 \times (10) ^2 =126\ m\$Net force \$ = 7 + (1.96) = 5.04\ N\$Work done by the net force, \$W_{net}= 5.04 \times 126 = 635\ J\$

Q: A \$(trolley + child)\$ of total mass \$200\ kg\$ is moving with a uniform speed of \$36\ km/h\$ on a frictionless track. The child of mass \$20\ kg\$ starts running on the trolley from one end to the other \$(10\ m\$ away) with the speed of \$10\ ms^{-1}\$ relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Mass of the trolley, \$ M=200kg\$Speed of the trolley, \$v = 36 km/h = 10 m/s\$Mass of the boy, \$m = 20 kg\$Initial momentum of the system of the boy and the trolley \$=(M + m)v\$ \$=(200 + 20) \times 10\$ \$=2200 kg \ m/s\$Let \$v'\$ be the final velocity of the trolley with respect to the ground.Final velocity of the boy with respect to the ground \$= v' - 4\$Final momentum \$=Mv' + m(v' - 4)\$ \$=200v' + 20v' - 80\$From the conservation of linear momentum:Initial momentum\$ =\$ Final momentum \$2200= 220 v'-80 \$ ∴ \$v' = 2280 / 220 = 10.36 m/s\$Length of the trolley, \$l= 10 m\$Speed of the boy, \$v''= 4 m/s\$ Time taken by the boy to run, \$t = 10/4 = 2.5 s\$∴ Distance moved by the trolley \$=v'' \times t\$= \$10.36\times 2.5 = 25.9m \$

Q: Answer the following :(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained. The rocket or the atmosphere?(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comets velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? (d) In Figure (i) the man walks \$2\$ \$m\$ carrying a mass of \$15\$ \$kg\$ on his hands. In Figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of \$15\$ \$kg\$ hangs at its other end. In which case is the work done greater?

(a) Rocket: The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket. According to the conservation of energy:Total Energy = Potential Energy + Kinetic Energy\$=mgh+(1/2)mv^2\$The reduction in the mass of the rocket causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket. (b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero. (c) When an artificial satellite, orbiting around the earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount. (d) Case(i)Mass, m = 15 kg Displacement, s = 2 mWork done, \$W = Fs cos\theta \$Where, \$\theta= Angle\, between\, force\, and\, displacement\$\$W=mgs cos\theta=15 \times 2 \times 9.8 \times cos90^0 = 0\$ Case(ii)Mass, m = 15 kg Displacement, s = 2 m Here, the direction of the force applied to the rope and the direction of the displacement of the rope are same. Therefore, the angle between them, \$= 0\$ Since \$cos \ 0^0 = 1\$Work done, \$W=Fscos\theta = mgs\$ \$= 15\times 9.8 \times 2 = 294 \ J\$ Hence, more work is done in the figure (ii).

Q: A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg \$s^{-1}\$. What is the speed of the trolley after the entire sand bag is empty ?

Since the hole is on the floor , that means sand is falling vertically with respect to trolley. Therefore there is no force in horizontal direction hence in horizontal direction momentum is conserved.Let \$M=\$ mass of trolley \$m=\$ mass of sandbag \$v_1=\$ initial velocity \$v_2=\$ final velocity ( to be found) Then \$P_1= (M+m)v_1\$ when the sand bag is empty the momentum is \$P_2= (M+0) v_2\$Momentum is conserved in horizontal direction so\$P_1= P_2\$\$\Rightarrow = v_2= \dfrac{(M+m)}{M} v_1= \dfrac{300+25}{300} \times 27 \times \dfrac{5}{18} = 8.215 m/s\$

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Work, Energy and Power

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Exercise

A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15min. If the tank is 40m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?[Dencity of water 103kgm−3](g=9.8ms−2)

A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s−1. What is the speed of the trolley after the entire sand bag is empty ?