AIPMT Questions (MCQ) from Work, Energy & Power

Here are two multiple choice questions from the section, ‘work, energy and power’. The first question appeared in AIPMT 2010 question paper while the second question appeared in AIPMT 2008 question paper.

(i) An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 ms^–1 .The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?

(1) 800 W

(2) 400 W

(3) 200 W

(4) 100 W

The power of the engine is equal to the kinetic energy of water flowing out per second. The mass (M) of water flowing out per second is 2×100 = 200 kg/s

Kinetic energy of water flowing out per second = ½ Mv² = ½ ×200×2² = 400 W.

(ii) Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10 % of energy. How much power is generated by the turbine? (g = 10 ms^–2)

(1) 7.0 kW

(2) 8.1 kW

(3) 10.2 kW

(4) 12.3 kW

The input power to the turbine is the potential energy due to the water falling per second and is equal to mgh = 15×10×60 =9000 watt.

The power generated by the turbine is 10 % less than the above value and is equal to (9000 – 900) watt = 8100 watt = 8.1 kW.

You will find similar useful questions in this section here as well as here.

Two Questions (EAMCET 2009 and EAMCET 2008) on Energy and Power

Today we will discuss two multiple choice questions involving energy and power. The first question appeared in the EAMCET 2009 (Engineering) question paper and the second question appeared in the EAMCET 2008 (Engineering) question paper:

(1) A motor of power P₀ is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to P₁. The ratio of P₁ to P₀ is

(1) n : 1

(2) n² : 1

(3) n³ : 1

(4) n⁴ : 1

If the mass of water delivered per second is m when the power is P₀, we have

P₀ α ½ mv² where v is the velocity of water

[We do not equate P₀ to ½ mv² since the efficiency of the motor will not be 100%].

To increase the rate of flow of water through the same pipe n times, the velocity of water has to be made n times. The mass of water delivered per second then is nm. Therefore we have

P₁ α ½ nm (nv)²

From the above we obtain

P₁/ P₀ = n³ [Option (3)].

(2) A river of salty water is flowing with a velocity 2 ms^–1. If the density of water is 1.2 g cm^–3, then the kinetic energy of each cubic metre of water is

(1) 2.4 J

(2) 24 J

(3) 2.4 kJ

(4) 4.8 kJ

The density of the water in the river is 1.2 g cm^–3, which is equal to 1200 kg m^–3. Mass of one cubic metre of the water is 1200 kg and its kinetic energy E is given by

E = ½ mv² = ½ ×1200×2² = 2400 J = 2.4 kJ.

EAMCET (Medical) 2009 Questions on Work and Energy

The following questions which appeared in the EAMCET (Medical) 2009 question paper are worth noting:

(1) A block of mass ‘m’ is connected to one end of a spring of spring constant ‘k’. The other end of the spring is fixed to a rigid support. If the mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then ‘d’ is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g = acceleration due to gravity)

(1) mg/k

(2) 2d

(3) mg/3k

(4) 4d

The mass m is suspended by means of the spring. Since the spring is extended through a distance d, we have

mg = kd so that k = mg/d

When the mass is suddenly released, suppose the spring extends through an additional distance x. The total extension then is d+x.

The spring mass system momentarily comes to rest in the condition of maximum extension and then tries to return to the initial extension of d, executing simple harmonic oscillations. In the condition of maximum extension (equal to d+x) the gravitational potential energy mg(d+x) of the mass is converted into elastic potential energy of the spring so that we have

mg(d+x) = (½) k(d+x)²

Or, mg(d+x) = ½ (mg/d)(d+x)² since k = mg/d

This gives 2 = (d+x)/d from which x = d

The total extension d+x is therefore equal to 2d [Option (2)]

(2) A particle is projected up from a point at an angle θ, with the horizontal direction. At any time ‘t’, if ‘p’ is its linear momentum, ‘y’ is the vertical displacement and ‘x’ is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy of the projectile is

(1) Graph (A)

(2) Graph (B)

(3) Graph (C)

(4) Graph (D)

The kinetic energy of a projectile has to decrease with the increase in its vertical displacement since its gravitational potential energy increases at the cost of its kinetic energy. Therefore graph (A) is incorrect.

[Graphs (B) and (C) are correct since the kinetic energy decreases with the increase in the horizontal displacement x, becomes a minimum at half the horizontal range (corresponding to the maximum height) and then increases. Graph (D) also is correct since the kinetic energy k is given by

k = p²/2m where p is the linear momentum and m is the mass of the particle.

Therefore, k is directly proportional to p², yielding a straight line graph].

Kerala Engineering Entrance 2009 Multiple Choice Questions on Work and Energy

In the KEAM (Engineering ) 2009 question paper three questions were included from the section ‘work, energy and power’. Here are those questions with solution:

(1) A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is

(a) 45 J

(b) 30 J

(c) 70 J

(d) 135 J

(e) 20 J

The work done by a variable force acting along the direction of displacement can be found by drawing a force-displacement graph. The area under this graph gives the work done. Since the force has positive and negative values, the area also has positive (+50 units) and negative (– 30 units) values. The net area is +20 units and hence the gain in kinnetic energy during the movement from position x = 0 to the position x = 16 m is 20 J.

The particle has kinetic energy of 25 J at position x = 0. Therefore the kinetic energy at position x = 16 m is 25 + 20 = 45 J.

(2) Two springs P and Q of force constants k_P and k_Q [k_Q = k_P/2] are stretched by applying force of equal magnitude. If the energy stored in in Q is E, then the energy stored in P is

(a) E

(b) 2 E

(c) E/8

(d) E/4

(e) E/2

The potential energy (U) of a spring stretched (by a force F) through distance x is given by

U = ½ kx² where k is the spring constant (force constant) given by k = F/x

This can be rewritten as

U = ½ (F²/k) since x = F/k

The nergy stored in P and Q are respectively given by U_P = ½ (F²/k_P) and U_Q = ½ (F²/k_Q).

Therefore, U_P/U_Q = k_Q/k_P = ½ as given in the question.

This gives U_P = U_Q/2 = E/2.

(3) A rod of mass m and length l is made to stand at an angle of 60º with the vertical. Potential energy of this rod in this position is

(a) mgl

(b) mgl/2

(c) mgl/3

(d) mgl/4

(e) mgl/√2

When the rod is kept inclined at an angle of 60º with the vertical, its centre of gravity is raised (from the ground level) by a height h = (l/2)cos 60º = l/4.

The gravitational potential energy of the rod in this position is mgh = mgl/4.

You will find many useful multiple choice questions on work, energy and power at physicsplus and at AP Physics Resources

Two Multiple Choice Questions involving Kinetic Energy

Check whether you can find the answers to the following two questions in a couple of minutes:

(1) When a running boy increases his speed by 2 ms^–1, his kinetic energy becomes three times the original value. His original speed is (in ms^–1)

(a) √3 +1

(b) √3 –1

(c) √3

(d) 2√3

(e) 2

If his original speed is ‘v’ we can write

(½)mv²×3 = (½)m(v+2)² where ‘m’ is his mass.

This gives v√3 = (v+2) so that v(√3 –1) =2.

Therefore, v = 2 /(√3 –1) = √3 +1 on multiplying the numerator and denominator by (√3 –1).

(2) A particle of mass ‘m’ at rest is acted on by a force ‘F’ for a time‘t’. Its kinetic energy after the time t is

(a) Ft²/2m

(b) F²t/2m

(c) F²t²/2m

(d) F²t²/3m

(e) F²t²/m

The impulse received by the particle during the time t is Ft. But impulse is equal to the change of momentum. Since the initial momentum of the particle is zero, Ft is the final momentum of the particle. The kinetic energy of the particle after the time t is therefore equal to F²t²/2m (remembering that KE = p²/2m where p is the momentum).

Two Multiple Choice Questions on Work and Energy

The following MCQ which appeared in KEAM (Engineering) 2007 question paper is worth noting:

A particle is released from a height S. At a certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

(a) S/4, 3gS/2 (b) S/4, [√(3gS)]/2 (c) S/2, [√(3gS)]/2

(d) S/4, √(3gS/2) (e) S/3, √(3gS/2)

Suppose the given conditions occur at the instant when the particle is at height ‘h’.

Potential energy at the height h = mgh.

Since the initial energy is completely potential and is equal to mgS, the kinetic energy at the height ‘h’ is equal to the loss of potential energy and is equal to mg(S–h).

Since the K.E. is three times the P.E., we have

mg(S–h) = 3mgh, from which h = S/4.

But, kinetic energy = ½ mv².

Therefore, ½ mv² = mg(S–h) = mg[S– (S/4)] = 3 mgS/4 so that

v =√(3gS/2)

Now, consider the following MCQ which is simple. But simple qiestions may cheat you some times. Here is the question:

A uniform chain of mass M and length L is placed on a table with 20% of its length overhanging from the edge of the table. The minimum work that has to be done to place the hanging portion on the table is

(a) MgL/5 (b) MgL/10 (c) MgL/20

(d) MgL/25 (e) MgL/50

The hanging portion has mass M/5 and therefore weight Mg/5. The length of the hanging portion is L/5 and hence the centre of gravity of the hanging portion is at a depth L/10 from the surface of the table. The entire weight of the hanging portion can be imagined to act through its centre of gravity. When you pull the hanging portion up to keep it on the table, you are moving the point of application of the force (equal to the magnitude of the the weight Mg/5) through a distance L/10 along the direction of the force.

Therefore, the work done is (Mg/5)×(L/10) = MgL/50

Two Kerala Medical Entrance 2007 Questions on Work and Energy

The following two questions appeared in Kerala Medical Entrance 2007 question paper which contained relatively simple questions:

(1) When a bullet is fired at a target, its velocity decreases by half after penetrating 30 cm in to it. The additional thickness it will penetrate before coming to rest is

(a) 30 cm (b) 40 cm (c) 10 cm (d) 50 cm (e) 20 cm

Questions similar to this one often find place in admission test question papers. This can be worked out using the work-energy principle or by using equations of one-dimensional motion. Let us use the work-energy principle:

Since the decrease in kinetic energy is equal to the work done against the retarding force in the target, we can write

½ m[v² – (v/2)²] = F×0.3, where ‘m’ is the mass of the bullet, ‘v’ is its initial velocity and F is the retarding force, which has to be assumed to be constant.

If ‘s’ is the additional distance penetrated (for the velocity to decrease to zero from the value v/2),

½ m(v/2)² = Fs

Dividing the first equation by the second, 3 = 0.3/s, from which s = 0.1 m = 10 cm.

[ If you use the equation of uniformly retarded linear motion, v² = u² –2as where ‘u’ and ‘v’ are the initial and final velocities respectively and ‘a’ is the retardation, we have the two equations,

(u/2)² = u² – 2a×0.3 and

0 = (u/2)² – 2as

The first equation is 3u²/4 = 2a×0.3.

The second equation is u²/4 = 2as.

Dividing the first one by the second, 3 = 0.3/s from which s = 01 m = 10 cm].

(2) A body constrained to move in the Y-direction is subjected to a force F = 2i + 15j + 6k newton. The work done by this force in moving the body through a distance of 10 m along the Y-axis is

(a) 100 J (b) ) 150 J (c) ) 120 J (d) ) 200 J (e) 50 J

The work (W) done by a force F in producing a displacement S is given by the scalar product of the vectors F and S.

Therefore, W = F.S = (2i + 15j + 6k) . 10j = 150 J

[The displacement vector is 10j since it is along the Y-axis].

You may work this out also by arguing that the component of the force along the direction of displacement of 10 m is 15 newton so that the work done is 15×10 = 150 J.

Spring Constant and Potential Energy of Spring

Spring constant (force constant of a spring) is the force required for unit extension (or contraction) in a spring. Suppose a spring is cut into two pieces of equal length. Will the spring constant change? Don’t be doubtful. The spring constant will be doubled. If you cut a spring of spring constant ‘k’ into ‘n’ pieces of equal length, the spring constant of each piece will be nk.
Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have springs of constants k₁, k₂, k₃….etc. in series, the net spring constant ‘k’ of the combination is given by the reciprocal relation, 1/k = 1/k₁ + 1/k₂ + 1/k₃ + . . . . . . .etc.
If you have springs in parallel, the net spring constant ‘k’of the combination is given by k = k₁+ k₂+ k₃+ . . . . .etc.
Often you will encounter questions involving the potential energy of a spring. You should remember that the potential energy of a spring of constant ‘k’, stretched (or contracted) through a distance ‘x’ is (½) kx². Consider the following MCQ:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be
(a) 3U (b) 6U (c) 9U (d) U (e) 36U
Since the potential energy is ½ kx², it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).
Suppose the above question is asked in the following modified form:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be
(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U
Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.
Now, consider the following question:
The tension in a spring of spring constant k is T. The potential energy of the spring is
(a) T²/k² (b) T²/k (c) 2T²/k (d) T²/2k (e) 2T²/k²
Potential energy of a spring, as you know, is (½)kx². But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.
Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is
(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k
When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx², from which x = 2Mg/k. [Option (b)]

Two Questions involving Energy and Power

(1) A bullet of mass 100g penetrates a distance of 10cm through a fixed block of mass 2kg. If the block is free to move, the distance penetrated will be nearly
(a) 3.5cm (b) 4.5cm (c) 5.5cm (d) 7.5cm (e) 9.5cm
If ‘F’ is the retarding force offered by the block and ‘s’ is the distance of penetration when the block is fixed, we have
Fs = ½ mu^2 ……………..(1)
where ‘m’ is the mass and ‘u’ is the initial velocity of the bullet. If ‘x’ is the distance of penetration when the block is free to move, we have
Fx = ½ mu^2 - ½ (M+m) v^2 where ‘M’ is the mass of the block and ‘v’ is the common velocity of the bullet and the block after the impact. Evidently, v = mu/(M+m) from the momentum conservation law so that we have
Fx = ½ mu^2 - ½ (M+m) [mu/(M+m)]^2
Simplifying, Fx = ½ Mmu^2 /(M+m) ………….(2)
Dividing eq(2) by eq(1) we get x/s = M/(M+m) so that x = Ms/(M+m) = 2×0.1/(2+0.1) = 0.095m = 9.5cm.
(2) An engine delivering constant power displaces a large block through 40cm in 4 seconds. What will be the displacement of the block in 16seconds?
(a) 1.6m (b) 3.2m (c) 4.8m (d) 6.4m (e) 8m
If the power is constant, the displacement will be proportional to t^(3/2) as shown below:
Power P = Fv = m(dv/dt)×v with usual notations. Since the power is constant (k) we have m(dv/dt)×v = k so that v(dv) = k(dt)/m. Integrating, we get
½ v^2 = kt/m. Therefore, v α√t. Since s=vt the displacement ‘s’ is proportional to t√t. Remember this: If power is constant, displacement is directly proportional to t^(3/2)
In the above problem therefore we can write
0.4/x = (4√4)/(16√16) =8/64, from which x = 3.2m