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]