`star` Introduction

`star` The Scalar Product

`star` The Work energy theorem

`star` The Scalar Product

`star` The Work energy theorem

`➢` The word ‘Work’ has a definite and precise meaning, it's define as measurement of energy transfer that occurs when an object is moved over a distance by an external force. Somebody who has the capacity to work for 14-16 hours a day is said to have a large stamina or energy.

`➢` Thus Energy is our capacity to do work. In Physics too, the term `color {blue}{"energy"}` is related to work in this sense, but as said above the term `color {blue}{"work"}` itself is defined much more precisely.

`➢` The word ‘power’ is used in everyday life with different shades of meaning.As per definition its the rate of doing work per unit time. e.g. In karate or boxing we talk of ‘powerful’ punches. These are delivered at a great speed. This shade of meaning is close to the meaning of the word `color {blue}{"power"}` used in physics.

`➢` Thus Energy is our capacity to do work. In Physics too, the term `color {blue}{"energy"}` is related to work in this sense, but as said above the term `color {blue}{"work"}` itself is defined much more precisely.

`➢` The word ‘power’ is used in everyday life with different shades of meaning.As per definition its the rate of doing work per unit time. e.g. In karate or boxing we talk of ‘powerful’ punches. These are delivered at a great speed. This shade of meaning is close to the meaning of the word `color {blue}{"power"}` used in physics.

`➢` There are two ways of multiplying vectors which we shall come across : one way known as the scalar product gives a scalar from two vectors and the other known as the vector product produces a new vector from two vectors.

`➢` Here we take up the scalar product of two vectors. The scalar product or dot product of any two vectors `A` and `B`, denoted as `A * B` (read `A` dot `B`) is defined as

`color {blue} {A * B = A B cos θ}`

`=>` where `θ` is the angle between the two vectors as shown in Fig.

`➢` Since `A, B` and `cos θ` are scalars, the dot product of `A` and `B` is a scalar quantity. Each vector, `A` and `B`, has a direction but their scalar product does not have a direction.

`➢` From Eq. (6.1a), we have `color {blue}{A*B = A (B cos θ )}`

`color {blue}{= B (A cos θ )}`

`color{red}☞` Geometrically, `B cos θ` is the projection of `B` onto `A` in Fig.6.1 (b) and `A cos θ ` is the projection of A onto B in Fig. 6.1 (c).

`➢` So, `A*B` is the product of the magnitude of `A` and the component of `B` along `A`. Alternatively, it is the product of the magnitude of `B` and the component of `A` along `B`.

`=>` Equation (6.1 a) shows that the scalar product follows the commutative law :

`A*B = B*A`

`color{red}☞` Scalar product obeys the distributive law:

`A* (B + C) = A*B + A*C`

Further, `A* (λ B) = λ (A*B)`

where `λ` is a real number.

`color{red}☞` For unit vectors `hat i , hat j , hat k` we have

`hat i * hat i = hat j * hat j = hat k * hat k=1`

`hat i * hat j = hat j * hat k = hat k * hat i =0`

`➢` Given two vectors

`A= A_x hat i + A_y hat j + A_z hat k`

`B= B_x hat i + B_y hat j + B_z hat k`

their scalar product is

`A * B = (A_x hat i + A_y hat j + A_z hat k ) * (B_x hat i + B_y hat j + B_z hat k)`

`color {blue}{= A_x B_x + A_y B_y + A_z B_z}` ..............(6.1b)

From the definition of scalar product and, (Eq. 6.1b) we have :

(i) `A * A = A_x A_x + A_y A_y + A_z A_z`

or, `color {blue}{A^2 = A_x^2 + A_y^2+ A_z^2}` ..............(6.1c)

since `A*A = |A ||A| cos 0 = A^2`.

(ii) `A*B = 0`, if `A` and `B` are perpendicular.

`➢` Here we take up the scalar product of two vectors. The scalar product or dot product of any two vectors `A` and `B`, denoted as `A * B` (read `A` dot `B`) is defined as

`color {blue} {A * B = A B cos θ}`

`=>` where `θ` is the angle between the two vectors as shown in Fig.

`➢` Since `A, B` and `cos θ` are scalars, the dot product of `A` and `B` is a scalar quantity. Each vector, `A` and `B`, has a direction but their scalar product does not have a direction.

`➢` From Eq. (6.1a), we have `color {blue}{A*B = A (B cos θ )}`

`color {blue}{= B (A cos θ )}`

`color{red}☞` Geometrically, `B cos θ` is the projection of `B` onto `A` in Fig.6.1 (b) and `A cos θ ` is the projection of A onto B in Fig. 6.1 (c).

`➢` So, `A*B` is the product of the magnitude of `A` and the component of `B` along `A`. Alternatively, it is the product of the magnitude of `B` and the component of `A` along `B`.

`=>` Equation (6.1 a) shows that the scalar product follows the commutative law :

`A*B = B*A`

`color{red}☞` Scalar product obeys the distributive law:

`A* (B + C) = A*B + A*C`

Further, `A* (λ B) = λ (A*B)`

where `λ` is a real number.

`color{red}☞` For unit vectors `hat i , hat j , hat k` we have

`hat i * hat i = hat j * hat j = hat k * hat k=1`

`hat i * hat j = hat j * hat k = hat k * hat i =0`

`➢` Given two vectors

`A= A_x hat i + A_y hat j + A_z hat k`

`B= B_x hat i + B_y hat j + B_z hat k`

their scalar product is

`A * B = (A_x hat i + A_y hat j + A_z hat k ) * (B_x hat i + B_y hat j + B_z hat k)`

`color {blue}{= A_x B_x + A_y B_y + A_z B_z}` ..............(6.1b)

From the definition of scalar product and, (Eq. 6.1b) we have :

(i) `A * A = A_x A_x + A_y A_y + A_z A_z`

or, `color {blue}{A^2 = A_x^2 + A_y^2+ A_z^2}` ..............(6.1c)

since `A*A = |A ||A| cos 0 = A^2`.

(ii) `A*B = 0`, if `A` and `B` are perpendicular.

Q 3149867713

Find the angle between force

`F = (3 hat i + 4 hat j + 5 hat k)` unit and displacement

`d = (5 hat i + 4 hat j + 3 hat k)` unit. Also find the projection of `F` on `d`.

`F = (3 hat i + 4 hat j + 5 hat k)` unit and displacement

`d = (5 hat i + 4 hat j + 3 hat k)` unit. Also find the projection of `F` on `d`.

`color{green} {F * d = F_x d_x + F_y d_y + F_z d_z}`

`= 3 (5) + 4 (4) + (– 3) (3)`

`= 16` unit

Hence `color{orange} {F*d = F d cosθ = 16}` unit

Now `color{purple} {F*F = F^2 =F_x^2+ F_y^2 = F _z^2}`

`= 9 + 16 + 25`

`= 50` unit

and `d*d = d^2 = d_x^2 +d_y^2 + d_z^2`

`= 25 + 16 + 9`

`= 50` unit

`∴ cos θ = 16/(sqrt 50 sqrt 50) = 16/50 = 0.32`

`➢` As we know, the relation for rectilinear motion under constant acceleration

`v^2 − u^2 = 2 as`

`=>` where `u` and `v` are the initial and final speeds and `s` the distance traversed. Multiplying both sides by `m//2`, we have

`color {blue}{1/2 mv^2 - 1/2 m u^2 = mas= Fs}` ..............(6.2a)

`=>` where the last step follows from Newton’s Second Law. We can generalise Eq. (6.1) to three dimensions by employing vectors

`v^2 − u^2 = 2 a*d`

`➢` Once again multiplying both sides by `m//2` , we obtain

`color {blue}{1/2 mv^2 - 1/2 mu^2 = m a * d = F* d}` ..................(6.2b)

`➢` The left side of the equation is the difference in the quantity ‘half the mass times the square of the speed’ from its initial value to its final value.

`➢` We call each of these quantities the ‘kinetic energy’, denoted by `K`. The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by `W`. Eq. (6.2) is then

`color {blue}{K_f - K_t = W}`.............(6.3)

`=>` where `K_i` and `K_f` are respectively the initial and final kinetic energies of the object. Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement.

`v^2 − u^2 = 2 as`

`=>` where `u` and `v` are the initial and final speeds and `s` the distance traversed. Multiplying both sides by `m//2`, we have

`color {blue}{1/2 mv^2 - 1/2 m u^2 = mas= Fs}` ..............(6.2a)

`=>` where the last step follows from Newton’s Second Law. We can generalise Eq. (6.1) to three dimensions by employing vectors

`v^2 − u^2 = 2 a*d`

`➢` Once again multiplying both sides by `m//2` , we obtain

`color {blue}{1/2 mv^2 - 1/2 mu^2 = m a * d = F* d}` ..................(6.2b)

`➢` The left side of the equation is the difference in the quantity ‘half the mass times the square of the speed’ from its initial value to its final value.

`➢` We call each of these quantities the ‘kinetic energy’, denoted by `K`. The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by `W`. Eq. (6.2) is then

`color {blue}{K_f - K_t = W}`.............(6.3)

`=>` where `K_i` and `K_f` are respectively the initial and final kinetic energies of the object. Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement.

Equation (6.2) is also known as `"work-energy (WE) theorem :"` The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section.

Q 3159067814

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined.

Consider a drop of mass `1.00 g` falling from a height 1.00 km. It hits the ground with a speed of `50.0 m s^(-1)`.

(a) What is the work done by the gravitational force ? What is the work done by the unknown resistive

force?

Consider a drop of mass `1.00 g` falling from a height 1.00 km. It hits the ground with a speed of `50.0 m s^(-1)`.

(a) What is the work done by the gravitational force ? What is the work done by the unknown resistive

force?

The change in kinetic energy of the drop is

`color{purple}{Delta K= 1/2 mv^2 -0}`

`=1/2 xx 10^(-3) xx 50 xx 50`

`1.25 J`

where we have assumed that the drop is initially at rest.

Assuming that `g` is a constant with a value `10 m//s^2`, the work done by the gravitational force

is,

`color{green}{W_g = mgh}`

`=10^(-3) xx 10 xx 10^3`

`=10.0 J`

(b) From the work-energy theorem

` color{orange}{ΔK = W_g +W_r}`

where `W_r` is the work done by the resistive force on the raindrop. Thus

`color{purple} {W_r = ΔK − W_g}`

`= 1.25 −10`