NEET 2013 Questions (MCQ) from Work and Energy

“The world is a dangerous place, not because of those who do evil, but because of those 

who look on and do nothing.” 

– Albert Einstein 

The following questions on work and energy appeared in the National Eligibility Cum Entrance Test (NEET) 2013 which replaced AIPMT for admitting students to MBBS and BDS courses:.

(1) A uniform force of (3 i + j) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position (2 i + k) metre to position (4 i + 3 j – k) metre. The work done by the force on the particle is

(1) 6 J

(2) 13 J

(3) 15 J

(4) 9 J

Work W is the scalar product (dot product) of force F and displacement s.

Or, W = F.s

Since the particle is displaced from position (2 i + k) metre to position (4 i + 3 j – k) metre, the displacement is given by

            s = (4 i + 3 j – k) – (2 i + k) = (2 i + 3 j – 2 k)

Therefore work W = F.s = (3 i + j) . (2 i + 3 j – 2 k) = 6 + 3 = 9 joule [Option (4)].

(2)The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

(1) μ = 2/tanθ

(2) μ = 2 tanθ

(3) μ = tanθ

(4) μ = 1/tanθ

This question is popular among question setters (See AIEEE 2005 question paper).

The component of gravitational force along the incline is mg sin θ where m is the mass of the object and g is the acceleration due to gravitaty.

The work done by the gravitational force when the object moves along the incline is mgL sinθ since the force mg sinθ acting along the incline moves the object through the length L of the incline.

The work done by the gravitational force imparts kinetic energy to the object. But the entire kinetic energy is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

            mgL sinθ = μmg cosθ (L/2)

[Note that the frictional force is μmg cosθ since the normal force exerted (by the incline) on the object is mg cosθ and the coefficient of friction is μ]

From the above equation we obtain μ = 2 sinθ/cosθ = 2 tanθ [Option (2)].

[You may argue in an equivalent manner like this too:

The object falls through a height L sinθ, thereby losing gravitational potential energy mgL sinθ. The entire gravitational potential is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

            mgL sinθ = μmg cosθ (L/2) from which μ = 2 tanθ

*          *          *          *          *          *          *          *          *          *          *          *

You can work out the above problem equally well by using the equation of linear motion,

            v² = u² + 2as

Considering the uniformly accelerated motion of the object down the upper smooth half of the inclined plane, the final velocity ‘v’ of the object is given by

            v² = 0 + 2 g sinθ (L/2) ………..(i)

Considering the uniformly retarded motion of the object down the lower rough half of the inclined plane, we have

            0 = v² + 2(g sinθ – μg cosθ) (L/2) ………(ii)

Substituting for v² from Eq. (i), we have

            2 gL sinθ = μgL cosθ, from which μ = 2 tanθ]

AIPMT 2009 Multiple Choice Questions on Work, Energy & Power

How strange is the lot of we mortals! Each of us is here for a brief sojourn; for what purpose we know not, though sometimes sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our own happiness depends.

–Albert Einstein

Three questions from the section ‘work, energy & power’ were included in the AIPMT 2009 question paper. Here are those questions with solution:

(1) A block of mass M is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

(1) 2 Mg/k

(2) 4 Mg/k

(3) Mg/2k

(4) Mg/k

If x is the maximum extension produced, we have

Mgx = ½ kx², on equating the decrease in the gravitational potential energy of the mass M to the increase in the elastic potential energy of the spring.

Therefore, x = 2Mg/k

(2) A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 ms^–2)

(1) 30 J

(2) 40 J

(3) 10 J

(4) 20 J

The initial kinetic energy of the 1 kg mass is given by

½ mv² = ½ ×1×20² = 200 J.

The gravitational potential energy of the 1 kg mass at its maximum height is given by

Mgh = 1×10×18 = 180 J.

The energy lost due to air friction is therefore equal to 200 – 180 = 20 J.

(3) An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

(1) mv²

(2) ½ mv²

(3) ½ m²v²

(4) ½ mv³

Mas of water flowing out per second through the hose is mv. Therefore, kinetic energy imparted per second to the water is ½ ×(mv)×v² = ½ mv³ [Option (3)].

You will find useful posts in this section here.

Kerala Engineering Entrance 2008 and other Multiple Choice Questions (MCQ) on Work and Energy

Setting an example is not the main means of influencing others; it is the only means.

– Albert Einstein

Let us discuss a few multiple choice questions on work and energy.

(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms^–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?

(a) 10 J

(b) –10 J

(c) – 1 J

(d) 1 J

(e) 9.8 J

When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv²) of the sphere but its sign is negative.

Therefore, the answer is – ½ mv² = – ½ ×0.020×10² = –1 J.

(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

(a) 2 × 10⁵ J to 5 × 10⁵ J

(b) 2 × 10⁴ J to 5 × 10⁴ J

(c) 2 × 10³ J to 5 × 10³ J

(d) 200 J to 500 J

(e) 20 J to 50 J

This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.

In the 100 m dash the velocity v is almost constant so that we have

v = 100 m/10 s = 10 ms^–1

We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.

His kinetic energy will be ½ mv² = ½ ×80×10² = 4000 J so that the correct option is (c).

Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.

(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t_A and t_B, then the ratio t_A/ t_B is

(a) ½

(b) 2

(c) 2/5

(d) 5/6

(e) 1/5

The concept of impulse will be very useful here. The impulse received by A and B in the times t_A and t_B are respectively F t_A and F t_B where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F t_A and F t_B.

Since the kinetic energy is p²/2m where p is the momentum, we have

(F t_A)²/2m_A = (F t_B)²/2m_B

Therefore, t_A/ t_B = √( m_A/ m_B) = √(20/5) = 2.

(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is

(a) 20

(b) 40

(c) 50

(d) 30

(e) 35

The resutant force (F) on the particle is the sum of the forces given by

F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.

The displacement (s) of the particle is given by

s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)

The work done (W) is given by

W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.

You will find similar useful multiple choice questions (with solution) at AP Physics Resources.

Multiple Choice Questions on Work and Energy

Here is a simple question which is meant for gauging your understanding of the work-energy principle. This MCQ appeared in AIEEE 2005 question paper:

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

(a) ML²/K (b) zero (c) KL²/2M (d) √(MK).L

If the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p²/2M. This must be equal to the maximum potential energy of the spring so that we have

p²/2M = ½ KL², from which p = L√(MK)., given in option (d).

Now consider the following MCQ which appeared in Kerala Medical Entrance 2006 question paper:

The work done by a force F = –6x³ î newton, in displacing a particle from x = 4m to x = –2m is

(a) 360 J (b) 240 J (c) – 240 J (d) – 360 J (e) 408 J

This is a case of variable force (in the X-direction), the point of application of which is moved in the X-direction. The work done is therefore given by

W = ∫F.dx = ₄ ∫^–2 (–6x³)dx = –6[x⁴/4] with x between limits 4 and –2.

Therefore, W = –(6/4)(16 – 256) = 360 joule, given in option (a).

The following MCQ appeared in Kerala Engineering entrance 2006 question paper:

A running man has the same kinetic energy as that of a boy of half the mass. The man speeds up by 2 ms^–1 and the boy changes his speed by ‘x’ ms^–1 so that the kinetic energies of the boy and the man are again equal. Then ‘x’ in ms^–1 is

(a) – 2√2 (b) + 2√2 (c) √2 (d) 2 (e) 1/√2

If the mass of the man is ‘m’, the mass of the boy is m/2. If v₁ and v₂ are the initial velocities of the man and boy respectively, we have

½ mv₁² = ½ (m/2)v₂²

Therefore, v₂ = v₁√2.

On changing the speeds, we have

½ m(v₁+2)² = ½ (m/2)(v₂+x)²

On substituting for v₂ (=v₁√2), the above equation simplifies to

(v₁+2)² = ½ (v₁√2+x)² from which x = 2√2 ms^–1, given in option (b).