Chapter 6 Work Energy and power Solutions
Question - 21 : - The blades of a windmill sweep out a circle of areaA. (a) If the wind flows at a velocity v perpendicular to the circle, what isthe mass of the air passing through it in time t? (b) What is the kineticenergy of the air? (c) Assume that the windmill converts 25% of the wind’senergy into electrical energy, and that A = 30 m2,v = 36 km/h and the density of air is 1.2 kg m-3 .What is the electrical power produced?
Answer - 21 : - (a) Volume of wind flowing per second = Av Mass ofwind flowing per second = Avρ
Mass of air passing in second = Avρt
Question - 22 : - A person trying to lose weight (dieter) lifts a 10kg mass, one thousand times, to a height of 0.5 m each time. Assume that thepotential energy lost each time she lowers the mass is dissipated, (a) How muchwork does she do against the gravitational force? (b) Fat supplies 3.8 x 107J of energy per kilogram which isconverted to mechanical energy with a 20% efficiency rate. How much fat willthe dieter use up?
Answer - 22 : - Here, m = 10 kg, h = 0.5 m, n = 1000
(a) work done against gravitational force.
W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000J.
(b) Mechanical energy supplied by 1 kg of fat = 3.8 x 107 x20/100
= 0.76 x107 J/kg
.-. Fat used up by the dieter =1kg/(0.76 x 107)x 49000 = 6.45 x 10-3 kg
Question - 23 : - A family uses 8 kW of power, (a) Direct solar energyis incident on the horizontal surface at an average rate of 200 W per squaremeter. If 20% of this energy can be converted to useful electrical energy, howlarge an area is needed to supply 8 kW? (b) Compare this area to that of theroof of a typical house.
Answer - 23 : - (a) Power used by family, p = 8 KW = 8000 W
As only 20% of solar energy can be converted to useful electrical energy,hence, power
8000 W to be supplied by solar energy = 8000 W/20 = 40000 W
As solar energy is incident at a rate of 200 Wm-2,hence the area needed
A=4000 W/200 Wm-2 =200 m2
(b) The area needed is camparable to roof area of a large sized house.
Question - 24 : - A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block of wood of mass0.4 kg and instantly comes to rest with respect to the block. The block issuspended from the ceiling by thin wire. Calculate the height to which theblock rises. Also, estimate the amount of heat produced in the block.
Answer - 24 : - Here, m1 =0.012 kg, u1 =70 m/s
m2 = 0.4 kg, u2 = 0
As the bullet comes to rest with respect to the block, the two behave as onebody. Let v be the velocity acquired by the combination.
Applying principle of conservation of linear momentum, (m1 + m2) v = m1H1 +m2u2 = m1u1
Question - 25 : - Two inclined frictionless tracks, one gradual andthe other steep meet at A from where two stones are allowed to slide down fromrest, one on each track (Fig). Will the stones reach the bottom at the sametime? Will they reach there at the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and h = 10m, what are the speeds and times taken by the two stones?
Answer - 25 : - 
Question - 26 : - A 1 kg block situated on a rough incline isconnected to a spring with spring constant 100 Nm-1 asshown in Figure. The block is released from rest with the spring in theunstretched position. The block moves 10 cm down the incline before coming torest. Find the coefficient of friction between the block and the incline.Assume that the spring has negligible mass and the pulley is frictionless.
Answer - 26 : - 
Question - 27 : - A bolt of mass 0.3 kg falls from the ceiling of anelevator moving down with a uniform speed of 7 ms-1.It hits the floor of the elevator (length of elevator = 3 m) and does notrebound. What is the heat produced by the impact? Would your answer bedifferent if the elevator were stationary?
Answer - 27 : - P.E. of bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J
The bolt does not rebound. So the whole of the energy is converted into heat.Since the value of acceleration due to gravity is the same in all inertialsystem, therefore the answer will not change even if the elevator isstationary.
Question - 28 : - A trolley of mass 200 kg moves with a uniform speedof 36 km h-1 on a friction lesstrack. A child of mass 20 kg runs on the trolley from one end to the other (10m away) with a speed of 4 ms-1 relativeto the trolley in a direction opposite to the trolley’s motion, and jumps outof the trolley. What is the final speed of the trolley? How much has thetrolley moved from the time the child begins to run?
Answer - 28 : - Let there be an observer travelling parallel to thetrolley with the same speed. He will observe the initial momentum of thetrolley of mass M and child of mass m as zero. When the child jumps in oppositedirection, he will observe the increase in the velocity of the trolley by Δv.
Let u be the velocity of the child. He will observe child landing at velocity(u – Δu) Therefore, initial momentum = 0
Final momentum = MΔ v – m (u – Δv)
Hence, MΔ v – m (u – Δv) = 0
Whence Δv =mu/ M + m
Putting values Δv =4 x 20/ 20 + 220 = ms-1
.-. Final speed of trolley is 10.36 ms-1.
The child take 2.5 s to run on the trolley.
Therefore, the trolley moves a distance = 2.5 x 10.36 m = 25.9 m
Question - 29 : - Which of the following potential energy curves inFig. cannot possibly describe the elastic collision of two billiard balls? Herer is distance between centres of the balls.
Answer - 29 : - The potential energy of a system of two masses variesinversely as the distance (r) between 1
them i.e., V (r) α 1/r. When the two billiard balls touch each other, P.E.becomes zero i.e., at r = R + R = 2 R; V (r) = 0. Out of the given graphs,curve (v) only satisfies these two conditions. Therefore, all other curvescannot possibly describe the elastic collision of two billiard balls.
Question - 30 : - Consider the decay of a free neutron at rest: n >p + e–. Show that the two body decayof this type must necessarily give an electron of fixed energy, andtherefore, cannot account for the observed continuous energy distribution inthe β -decay of a neutron or a nucleus, Fig.
Answer - 30 : - Let the masses of the electron and proton be m and Mrespectively. Let v and V be the velocities of electron and protonrespectively. Using law of conservation of momentum. Momentum of electron +momentum of proton = momentum of neutron
Thus, it is proved that the value of v2 isfixed since all the quantities in right hand side are constant. It establishesthat the emitted electron must have a fixed energy and thus we cannot accountfor the continuous energy distribution in the β-decay of a neutron.