Topic covered

`color{red} star` WORK
`color{red} star` KINETIC ENERGY


`➢` As seen earlier, work is related to force and the displacement over which it acts.
`➢` Let's Consider a constant force `F` acting on an object of mass `m`. The object undergoes a displacement d in the positive x-direction as shown in Fig. 6.2.

`➢` The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus

`color {blue}{W= (F cos theta)d= F*d}`


`➢` `"We see that if there is no displacement, there is no work done even if the force is large."`
`➢` Thus, when you push hard against a rigid brick wall, the force you exert on the wall does no work. Yet your muscles are alternatively contracting and relaxing and internal energy is being used up and you do get tired.
`➢` So here, the meaning of work in physics is different from its usage in everyday language.

`color{red}bbul{"No work is done if :"}`

(i) the displacement is zero as seen in the example above. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time.

(ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement.

(iii) the force and displacement are mutually perpendicular.
`=>` This is so since, for `θ = π//2` rad (`= 90^o), cos (π //2) = 0`. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement.
`=>` If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and `θ = π//2`.

`color{red}☞` Work can be both positive and negative. If `θ` is between `0^o` and `90^o, cos θ` in Eq. (6.4) is positive.

`➢` If `θ` is between `90^o` and `180^o, cos θ` is negative. In many examples the frictional force opposes displacement and `θ = 180^o`. Then the work done by friction is negative `(cos 180^o = –1)`.

`●` From Eq. (6.4) it is clear that work and energy have the same dimensions, `[ML^2T^(–2)]`.
`●` The SI unit of these is joule (J), named after the famous British physicist James Prescott Joule (1811-1869). Since work and energy are so widely used as physical concepts, alternative units abound and some of these are listed in Table .

Alternative Units of Work/Energy in `J`

Q 3139167912

A cyclist comes to a skidding top in `10` m. During this process, the force on the cycle due to the road is 200 N and
is directly opposed to the motion.
(a) How much work does the road do on the cycle ?
(b) How much work does the cycle do on the road ?


Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.

(a) The stopping force and the displacement make an angle of `180^o` (`π` rad) with each other.

Thus, work done by the road,

`color{green} {W_r = Fd cosθ}`

`= 200 xx 10 xx cos π`

`= – 2000 J`

It is this negative work that brings the cycle to a halt in accordance with WE theorem.

(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is `200` N. However, the
road undergoes no displacement. Thus, work done by cycle on the road is zero.

The lesson of this example is that though the force on a body A exerted by the body B is always equal and opposite to that on B by A (Newton’s Third Law); the work done on `A` by `B` is not necessarily equal and opposite to the work done on `B` by `A`.


`➢` As we know, if an object of mass `m` has velocity `v`, its kinetic energy `K` is

`color {blue}{K= 1/2 m v* v = 1/2 mv^2}`


`➢` Kinetic energy is a scalar quantity. The kinetic energy of an object is a measure of the work an

Table 6.2 `color{blue}{"Typical kinetic energies (K)"}`

`=>` object can do by the virtue of its motion. This notion has been intuitively known for a long time.

`➢` The kinetic energy of a fast flowing stream has been used to grind corn. Sailing ships employ the kinetic energy of the wind. Table 6.2 lists the kinetic energies for various objects.
Q 3169167915

In a ballistics demonstration a police officer fires a bullet of mass `50.0 g` with speed `200 m s^(-1)` (see Table 6.2) on soft
plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the
bullet ?


The initial kinetic energy of the bullet is `mv^2//2 = 1000 J`. It has a final kinetic energy of `0.1 xx 1000 = 100 J`. If `v_f` is the emergent speed
of the bullet,

`color{purple} {1/2 mv_f^2 = 100J}`

`v_f= sqrt((2 xx 100 J)/ (0.05 kg))`

`= 63.2 ms^(-1)`

The speed is reduced by approximately `68%` (not `90%`).


`➢` constant force is rare. It is the variable force, which is more commonly encountered. Fig. 6.2 is a plot of a varying force in one dimension.

`➢` If the displacement `Δx` is small, we can take the force `F (x)` as approximately constant and the work done is then

`color{green} {ΔW =F (x) Δx}`

`=>` This is illustrated in Fig. 6.3(a). Adding successive rectangular areas in Fig. 6.3(a) we get the total work done as

`color {blue}{W equiv sum_(x_i)^(x_f) F(x) Delta x}` .................(6.6)

`=>` where the summation is from the initial position `x_i` to the final position `x_f`.

`➢` If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 6.3(b). Then the work done is

`W= lim_(Delta x ->0) sum_(x_i)^(x_f) F (x) Delta x`

`color {green}{= int_(x_i)^(x_f) F(x) dx}`


`=>` where ‘`"lim"`’ stands for the limit of the sum when `Δx` tends to zero.
`=>` Thus, for a varying force the work done can be expressed as a definite integral of force over displacement (see also Appendix 3.1).
Q 3109578418

A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over
a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The
total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force,
which is 50 N. Calculate the work done by the two forces over 20 m.


The plot of the applied force is shown in Fig. 6.4. At x = 20 m, F = 50 N (≠ 0). We are given that the frictional force f is |f|= 50 N. It opposes
motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis.

The work done by the woman is

` color{green} {W_F →} color{purple} {" area of the rectangle ABCD + area of the trapezium CEID"}`

`W_F = 100 xx 10 + 1/2 (100 +50) xx 10`

`=1000 + 750`

`=1750 J`

The work done by the frictional force is

`color{green} {W_F →} ` area of the rectangle AGHI

`W_f = (−50) xx 20`

`= − 1000 J`

The area on the negative side of the force axis has a negative sign.