`“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 =

*μ*

*mg*cosθ (

*L/*2)

[Note that the frictional force is μ

*mg*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 =

*μ*

*mg*cosθ (

*L/*2) from which μ = 2 tanθ

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

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

v

^{2}= u^{2}+ 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

v

^{2}= 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
= v

^{2}+ 2(*g*sinθ – μ*g*cosθ) (*L/*2) ………(ii)
Substituting for v

^{2}from Eq. (i), we have
2

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