Work Energy Power
NCERT Solution
Part 2
Question 11: A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
F=-î +2ĵ +3k̂ N
Where, î, ĵ, k̂ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Answer: Work = Force × displacement
=(-î + 2ĵ + 3k̂ N) × 4m
= (0 + 0 + 3) × 4 = 3 × 4 = 12 J
Question 12: An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (Electron mass = 9.11 × 10-31 kg, proton mass = 1.67 × 10-27 kg, 1 eV = 1.60 × 10-19 J).
Answer: Given, Ke = 10 keV, Kp = 100 keV, me = 9.11 × 10-31 kg, and mp = 1.67 × 10-27 J
`K=1/2mv^2`
Or, `v^2=(2K)/m`
Or, `v=sqrt((2K)/m)`
Ratio of velocities of electron and proton can now be calculated as follows:
`(v_e)/(v_p)=( sqrt((2K_e)/m_e))/( sqrt((2K_p)/m_p))`
`=sqrt((K_e)/(m_e)xx(m_p)/(K_p))`
`=sqrt((10)/(100)xx(1.67xx10^(-27))/(9.11xx10^(-31)))`
`sqrt((1.67xx10^3)/(9.11))`
`=sqrt((1670)/(9.11))=sqrt(183)`
= 13 (approx)
Here, numerator is > denominator. So, speed of electron is more than that of proton.
Question 13: A rain drop of radius 2 m falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. 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 peed on reaching the ground is 10 m s-1?
Answer: First Step: Calculation of mass of raindrop
Given, r = 2 mm = 2 × 10-3 m
Distance covered in each half of journey = S = 500/2 = 250 m
Density of water = 10^3 kg m-3
Mass = Volume × density
`=4/3πr^3×ρ`
`=4/3xx(22)/7xx(2xx10^(-3))^3xx10^3`
`=4/3xx(22)/7xx8xx10^(-6)`
`=(704)/(21)xx10^(-6)=3.35xx10^(-5)` kg
No matter whether the drop moves with acceleration or with uniform speed, the work done by gravitational force remains the same.
In absence of resistive force, energy of drop on reaching the ground
`E_1=mg\h`
`=3.35xx10^(-5)xx9.8xx500=0.164` J
Actual energy because of presence of resistive force
`E_2=1/2mv^2`
`=1/2xx3.35xx10^(-5)10^2`
`=1.675xx10^(-3)=0.001675` J
Work done by resistive forces
`W = E_1-E_2`
`=0.164-0.001675=0.162` J
Question 14: A molecule in a gas container hits a horizontal wall with speed 200 m s-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Answer: The molecule rebounds with same speed at which it strikes the wall. So, momentum of molecule does not change. It means the collision is elastic. Wall remains at rest because its mass is significantly larger than mass of molecule.
Question 15: A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 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?
Answer: Calculation of mass of water
Mass = volume × density
= 30 × 103 kg
Work done = mgh
= 30 × 103 × 9.8 × 40
Now, time = 15 min = 15 × 60 = 900 s
Power = W/t
`=(30xx10^3xx9.8xx40)/(900)`
`=(39.2)/3xx10^3=13` W
As efficiency of pump is 30% so total power consumption
`(13xx100)/(30)=43.33` W
Question 16: Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following is a possible result after collision?
Answer: Let assume mass of each ball is m
Kinetic Energy before collision
Ball 1 `= 1/2mv^2`
Ball (2 + 3) = 0 (since balls are at rest)
So, KE of system of 3 balls before collision = `1/2mv^2`
Case 1: KE after collision `=0+1/2xx2mxx(v/2)^2`
`=1/4mv^2`
KE has changed in this case, i.e. was not conserved after collision.
Case 2: KE after collision `=1/2mv^2`
KE is conserved after collision.
Case 3: KE after collision `=1/2xx3mxx(v/3)^2`
`=1/6mv^2`
KE is not conserved in this case.
It is clear that case ii is the possible result.
Question 17: The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in this figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer: After collision, bob A will come to rest, while bob B will come into motion, because in case of elastic collision of two objects of same mass there is interchange of velocities.
Question 18: The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
Answer: Given, length l = 1.5 m, mass = m, dissipated energy = 5%
At horizontal position
Potential energy Ep = `mg\l`
Kinetic energy Ek = 0
So, total E = `mg\l`
At lowermost point (or median)
Ep = 0 and Ek = `1/2mv^2`
So, total E `=1/2mv^2`
As the bob loses 5% energy while coming down, it will be left with 95% of its total energy once it reaches the median position. This means
`1/2mv^2=0.95mg\l`
Or, `v^2=2xx0.95xx\gl`
`=2xx0.95xx9.8xx1.5=27.93`
Or, `v=sqrt(27.93)=5.28` m/s
Question 19: 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?
Answer: The fact that trolley is moving with uniform speed tells that no external force is working on it. This also means that no external force is involved in emptying the sand bad. So, trolley will continue to move at same speed even the sand bag becomes empty.
Question 20: A body of mass 0.5 kg travels in a straight line with velocity `v=ax^(3/2)` where `a=5m^(-1/2)` s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Answer: Initial Ek = 0 because v = 0 for x = 0
For x = 2, we get v = 5 × `2^(3/2)`
Final Ek `=1/2mv^2`
`=1/2xx0.5xx(5xx2^(3/2))^2`
`=0.25xx25xx2^3=0.25xx25xx8=50` J
Work done is 50 J as this is also the increase in KE.