Chapter 6 Work Energy and power Solutions
Question - 11 : - A body constrained to move along the z-axis of acoordinate system is subject to a constant force F given by
where i, j, k, are unit vectors along the x- y- andz-axis of the system respectively. What is the work done by this force inmoving the body a distance of 4 m along the z-axis?
Answer - 11 : - 
Question - 12 : - An electron and a proton are detected in a cosmicray experiment, the first with kinetic energy 10 keV, and the second with 100keV. Which is faster, the electron or the proton? Obtain the ratio of theirspeeds, (electron mass = 9.11 x 10-31 kg,proton mass = 1.67 x 10-27 kg, 1eV= 1.60 x 1019J).
Answer - 12 : - 
Question - 13 : - A raindrop of radius 2 mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscousresistance of the air) until at half its original height, it attains itsmaximum (terminal) speed, and moves with uniform speed thereafter. What is thework done by the gravitational force on the drop in the first and second halfof its journey? What is the work done by the resistive force in the entirejourney if its speed on reaching the ground is 10 ms-1?
Answer - 13 : -
Here, r = 2 mm = 2 x 10-3 m.
Distance moved in each half of the journey, S=500/2= 250 m.
Density of water, p = 103 kg/ m3
Mass of rain drop = volume of drop x density
m =4/3 π r2 x ρ =4/3 x 22/7 (2 x10-3)3 x103 = 3.35 x 10-5 kg
.-. W = mg x s = 3.35 x 10-5 x9.8 x 250 = 0.082 J
Note: Whether the drop moves with decreasing acceleration or with uniformspeed, work
done by the gravitational force on the drop remains the same.
If there was no resistive forces, energy of drop on reaching the ground.
E1= mgh = 3.35 x 10-5 x 9.8 x 500 = 0.164 J
Actual energy, E2 =1/2mv2 = 1/2 x 3.35 x 10-5 (10)2 =1.675 x 10-3J
Work done by the resistive forces, W =E1 – E2 =0.164 – 1.675 x 10-3 W
= 0.1623 joule.
Question - 14 : - A molecule in a gas container hits a horizontal wallwith speed 200 msA and angle 30° with the normal, and rebounds with the samespeed. Is momentum conserved in the collision? Is the collision elastic orinelastic?
Answer - 14 : - Let us consider the mass of the molecule be m and thatof wall be M. The wall remains at rest due to its large mass. Resolvingmomentum of the molecule along x-axis and y-axis, we get
The x-component of momentum of molecule
Question - 15 : - A pump on the ground floor of a building can pump upwater to fill a tank of volume 30 m3 in15 min. If the tank is 40 m above the ground, and the efficiency of the pump is30%, how much electric power is consumed by the pump?
Answer - 15 : -
Here, volume of water = 30 m3 ;t = 15 min = 15 x 60 = 900s
h = 40 m ; n= 30%
As the density of water = p = 103 kgm-3
Mass of water pumped, m = volume x density = 30 x 103 kg
Actual power consumed or output power p0 = W/t = mgh/t
=>p0=(30 x 103 x 9.8 x 40)/900=13070 watt
If pi is input power (required), then as
η=p0/pi=> pi=p0/η = 13070/(30/100)=43567 W =43.56 KW
Question - 16 : - Two identical ball bearings in contact with eachother and resting on a friction less table are hit head-on by another ballbearing of the same mass moving initially with a speed V. If the collision iselastic, which of the following (Fig.) is a possible result after collision?
Answer - 16 : - Let m be the mass of each ball bearing. Beforecollision, total K.E. of the system
=1/2mv2 + 0 =1/2 mv2
After collision, K.E. of the system is
Case I, E1 = 1/2(2m) (v/2)2 = 1/4 mv2
Case II, E2 =1/2 mv2
Case III, E3 =1/2(3m) (v/3)2 = 1/6mv2
Thus, case II is the only possibility since K.E. is conserved in this case.
Question - 17 : - The bob A of a pendulumreleased from 30° to the vertical hits another bob B of the same mass at reston a table as shown in Fig. How high does the bob A rise after the collision?Neglect the size of the bobs and assume the collision to be elastic.
Answer - 17 : -
Since collision is elastic therefore A would come to rest and Bwould begin to move with the velocity of A..
The bob transfers its entire momentum to the ball on the table.The bob does not rise at all.
Question - 18 : - The bob of a pendulum is released from a horizontalposition. If the length of the pendulum is 1.5 m, what is the speed with whichthe bob arrives at the lowermost point, given that it dissipated 5% of itsinitial energy against air resistance?
Answer - 18 : - On releasing the bob of pendulum from horizontalposition, it falls vertically downward by a distance equal to length ofpendulum i.e., h = l = 1.5 m .
As 5% of loss in P.E. is dissipated against air resistance, the balance 95%energy is transformed into K.E. Hence,
Question - 19 : - A trolley of mass 300 kg carrying a sandbag of 25 kgis moving uniformly with a speed of 27 km/h on a friction less track. After awhile, sand starts leaking out of a hole on the trolley’s floor at the rate of0.05 kg s-1. What is the speed of thetrolley after the entire sand bag is empty?
Answer - 19 : -
The system of trolley and sandbag is moving with a uniformspeed. Clearly, the system is not being acted upon by an external force. If thesand leaks out, even then no external force acts. So there shall be no changein the speed of the trolley.
Question - 20 : - A particle of mass 0.5 kg travels in a straight linewith velocity u = a x3/2, wherea = 5 m-1/2 s-1. What is the work done by the net forceduring its displacement from x = 0 to x = 2 m?
Answer - 20 : -
Here m = 0.5 kg
u=a x3/2, a = 5 m-1/2 s-1.
Initial velocity at x = 0, v1 =a x 0 = 0
Final velocity at x = 2, v2 =a (2)3/2 = 5 x (2)3/2
Work done = increase in K.E
= 1/2 m(v22-v12) = 1/2 x0.5[(5 x (2)3/2)2 – 0] = 50 J.