Education Journalist | Study Abroad Lead
Work, Energy and Power are the foundation of dynamics. Work can be defined as the displacement of the object when a unit force is applied to an object. It is a scalar quantity. Therefore the formula for work done is the product of the force vector and the displacement vector. Energy can be defined as the capacity to do work and it can neither be created nor be destroyed. Energy can be categorized into two forms- kinetic energy and potential energy. Power is defined as the rate of doing work or the amount of work done per unit of time. Power is also a scalar quantity. Power can be written as the ratio of work done divided by the time taken.
Work, Energy & Power
Very Short Answer Questions (1 Mark Questions)
Ques. If the two bodies stick after collision then what kind of collision is this- elastic or inelastic?
Ans. If two bodies stick together after colliding with each other then the collision is said to be an inelastic collision.
Read More: Elastic Collision
Ques. What is the triple point of water?
Ans. The triple point of water is the point where the pressure and temperature are such that water can co-exist in all three states of matter.
Ques. Defines dulong and Petit law.
Ans. Dulong and Petit’s law states that the value of specific heat of all the solids is constant at room temperature equal to 3R, where R is the universal gas constant.
Ques. Why only mercury is preferred for making thermometers?
Ans. Mercury is preferred in making thermometers as it has a wide range of temperatures where it is in liquid form and it has a uniform expansion rate.
Ques. Why are clock pendulums made of invar?
Ans. The clock pendulums are made of invar because it has a low value of the coefficient of linear expansion. So there will be a small change in the length of the pendulum for a small change in the temperature.
Ques. If the glass expanded more with increasing temperature than mercury, explain how would a thermometer be different?
Ans. The scale of the thermometer would be upside down if glass expanded more with increasing temperature than mercury.
Ques. When an arrow is shot, from where the arrow gets its KE?
Ans. It is the potential energy of the bent bow which is transferred to the KE of the arrow.
Read More: Thermal Properties of Matter
Short Answer Questions (2 Marks Questions)
Ques. A body is only allowed to move on the z-axis of a coordinate system. The body is subjected to a constant force F = -i + 2j + 3k N. Calculate the work done to move the body a distance of 4m along the z-axis.
Ans. Given the force F = -i + 2j + 3k N
Displacement = 4k m (where k is unit vector along z-axis)
Work done = Force * displacement
W = F.s
= ( -i + 2j + 3k) .(4k)
= 12 J
The work done to move the body is 12 J
Ques. A ball in a container hits a horizontal wall with some speed making an angle of 60° at normal, and rebounds with the same speed. Can we say that momentum is conserved in the collision? Is it an elastic or inelastic collision?
Ans. Yes, the collision is elastic. The momentum of the ball is conserved independent of the collision being elastic or inelastic. The ball has some velocity v and strikes the stationary wall of the container and rebounds with the same speed v. Therefore, the total kinetic energy of the ball is conserved during the collision.
Ques. The bob X of a pendulum released at an angle of 60° to the vertical hits another bob Y of the same mass at rest on a table. How high will the bob X rise after the collision? The radius of the bob is negligible and the collision is elastic.
Ans. Bob X will not rise after the collision. In an elastic collision between two objects with equal masses in which one is at rest, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass will immediately come to rest after the collision. In this case, a complete transfer of momentum takes place from the moving object to the stationary object. Thus, bob X of mass m will come to rest after colliding with bob Y, while bob Y will move with the velocity of bob X after the collision.
Ques. A trolley with a mass of 300 kg contains a sandbag of 25 kg. It is moving uniformly with a speed of 27 km/h on a frictionless track. After some interval of time, the sand begins leaking out of a hole on the floor of the trolley, at the rate of 0.05 kg/s. What is the speed of the trolley after the sandbag is empty?
Ans. The sandbag is placed on a trolley which is moving with a uniform speed of 27 km/h. The external forces acting on the whole system, which is (sandbag + trolley) is zero. As the leaking of the sand does not produce any external force on the system, the sand starts leaking from the bag with no change in the velocity of the trolley. Thus, the speed of the trolley will remain as it was before.
Read More: Difference between Speed and Velocity
Ques. Two bodies are at different temperatures Ta and Tb. They are brought in thermal contact but they do not necessarily settle down to the mean temperature. Why?
Ans. When two bodies at different temperatures Ta and Tb come in contact with each other, they do not always settle at their mean temperature because they might have different thermal capacities.
Ques. Calculate the work done by the centripetal force?
Ans. The work done by the centripetal force is always zero. This is because the angle, centripetal force makes with the displacement vector is always 90 degrees. Therefore the dot product is zero and hence work done is 0.
Long Answer Questions (3 Marks Questions)
Ques. A ball by mistake fell from a table on the floor. The height of the table is 2cm. It rises up to a height of 1.5m after its collision with the ground. Assume that 40% of mechanical energy is lost and it goes to thermal energy into the ball. Calculate the increase in temperature of the ball during the collision. Specific heat capacity of the ball = 800J/k. g = 10m/s2
Ans. Given in the question (initial height) h1 = 2m
(Final height) h2 = 1.5m
It is known that potential energy(PE) = mechanical energy for a body at rest (as kinetic energy is 0 )
ME lost = |mg(h1-h2)|
= |1 x 10 (105 -2 )|
= 5 J
Now,
ME x 40% = Cm\(\bigtriangleup\)T
40/100 x 5 = 800 x 1 x T
Therefore,
\(\bigtriangleup\)T = 100 x 800 /(40 x 5) = 2.5 x 10-3 oC .
This is the increase in the temperature.
Read More: Conservation Of Mechanical Energy
Ques. A thermometer is found to have the wrong calibration. The melting point of ice is read by it as 100oC. It reads 600oC in place of 500oC. What temperature will it show for the boiling point of water?
Ans. Given to us,
The lower fixed point on the wrong scale is 100oC
Let n be athe number of divisions between upper and lower fixed points on the scale
Let Q = reading on this scale,
(C−0)/100=Q−(−10)/n
Now, C = wrong reading = 600oC
Q = correct reading = 500oC
So,
(50−0)/100 = (60−(−10))/n
50/100=70/n
n = 140
Now,
C−0100=Q−(−10)n
Boiling point of water on the Celsius scale is 100oC
So, (100−0)/100 = (Q+10)/140
Q = 130oC,
This is the temperature of the boiling point of water on this scale.
Read More: Boiling Point Elevation
Ques. A body is moving in a single direction under the influence of the source of constant power. Its displacement in time t is proportional to
(i) t1/2
(ii) t
(iii) t3/2
(iv) t2
Ans. The correct option is option (iii) t3/2
Power can be written as,
P = Fv
= mav = mv dv/dt = k (constant)
⇒vdv=k/m dt
Integrating both the sides
v2/2=k/m x t
v=\(\sqrt (2kt/m)\)
Displacement of the body,
Dx = k′ t1/2dt
Where k′ =\(\sqrt (2k/3) \)
Integrating both sides,
X = 2/3k′ t3/2
⇒x∝t3/2
Conservation of Energy
Read More: The Principle of Conservation of Energy
Very Long Answer Questions (5 Marks Questions)
Ques. A bullet that has a mass of 0.012 kg and Vx = 70 ms-1 hits a block of wood of mass 0.4 kg and instantly comes to rest w.r.t the block. The block is suspended from the ceiling by a thin string. Calculate the height upto which the block will rise. Calculate the amount of heat produced.
Ans. Given to us that m1 = 0.012 kg ,u1 = 70m/s
m2 = 0.4 kg = 400g and u2 = 0m/s
The bullet and wood behave as one body because the bullet comes to rest with respect to the wood. Let v be the final velocity gained by the system.
As per the principle of conservation of linear momentum,
So, v = m1u1/(m1 + m2) = 0.012 x 70 /(0.012 + 0.4) = 0.84 / 0.412 = 2.04 m/s.
Let us assume that the block will rise up to a height h.
Total P.E = total K.E
(m1 + m2)gh = ½ (m1 + m2)v2
H = v2/2g
= 2.04 x 2.04 /(2 x 9.8) = 0.212m
The energy lost in the process is actually the heat lost.
Therefore Work done = initial energy - final energy
W = ½ m1 u12 - ½ (m1 + m2) v2
= ½ X 0.012 (70)2 - ½ X 0.412 (2.04)2
= 29.4 - 0.86
= 28.54 J
Heat lost = W/J = 28.54 / 4.2 = 6.8 cal.
The heat produced will be 6.8 cal.
Ques. A 1 kg block is placed on a rough inclined plane. It is connected to a spring having a spring constant of 100 N/m as shown in the diagram. The block is released from rest with the spring in a relaxed position. The block moves about 10 cm down the inclined plane before it comes to rest. Find (coefficient of friction) between the block and the inclined plane. Assume the spring has negligible mass and the pulleys are frictionless.
Ans. From the above figure,
Applying NLM of 1kg block
R = mg cos 37°
f= μR = mg sin 37°
Net force on the block down the incline = mg sin 37° – f
= mgsin 37° – μmgcos 37°
= mg(sin 37° – μcos 37°)
The distance moved by the object = 10cm = 0.1m
Let the distance moved to be x
In the equilibrium position, Work done = P.E of spring
Read More: Difference Between Momentum and Inertia
Ques. A raindrop having a radius of 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration due to air drag, until it reaches half its original height. It then attains its maximum speed and moves with uniform speed after that. Calculate what is the work done by the gravitational force on the drop in the 1st and 2nd half of its journey. What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10m/s?
Ans. Given to us
Radius of the drop, r = 2 mm = 2×10−3m
Density of water is known to us, ρ=103 kg/m3
Volume of the raindrop, V = 4/3 \(\pi \) r3
= 4/3 × 3.14 × (2×10−3)3 × 103 kg
Mass of the drop, m=ρ x V
=4/3×3.14×(2×10−3)3×103 kg
F = mg (gravitational force)
=4/3×3.14×(2×10−3)3×103 kg × 9.8N
The work done by the gravitational force on the drop in the first half,
W1= F. s
=4/3×3.14×(2×10−3)3×103 kg×9.8×250
= 0.082 J
This work done is mathematically equal to the work done by the gravitational force on the drop in the second half of its journey
W2 = 0.082 J
The law of conservation of energy states that if no external force is present, the total energy of the system remains the same.
Total energy at the top,
ET = mgh + 0
=4/3×3.14×(2×10−3)3×103 kg×9.8×500×10−5
=0.164J
Due to the presence of air drag (resistive force), the drop hits the ground with a velocity of 10 m/s.
Total energy at the ground will be:
EG = 12m = 4/3×3.14×(2×10−3)3×103 kg×9.8×102
= 1.675×10−3 Joules
Wr (Work done by the resistive force) = EG − ET = −0.162J
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