- #1

Pushoam

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Moved from a technical forum, so homework template missing

Question: A) Derive the work - energy theorem for one particle.

B) Check whether it is applicable for a system of particles and a rigid body

total sum of work done by individual forces = work done by total force

To show the above equality,

let's say that there acts n forces on a particle and the net displacement of the particle is from ## \vec s_i ## to ## \vec s_f##.

Then , total sum of work done by individual forces = ## Σ_{i =1} ^n W_i = Σ_{i =1} ^n \int _ {\vec s_i } ^ {\vec s_f} \vec F_i . d\, \vec s ##

## = \int _ {\vec s_i } ^ {\vec s_f} \{Σ_{i =1} ^n \vec F_i\} . d\, \vec s ##

## \vec F_{net} = Σ_{i =1} ^n \vec F_i ##

So, total sum of work done by individual forces = ## \int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s ## = work done by total force

Now, the work energy theorem says ,

The total work done on a particle by net force is equal to change in its kinetic energy provided that the mass of the particle remains constant.

## W = \int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s = \int _ {t_i } ^ {t_f} m \frac {d \vec v } {dt} . \vec v d\, t ##

## = \int _ {v_i } ^ {v_f} m \frac 1 2 d\, v^2 = ## ## \frac 1 2 m v_f ^2 - \frac 1 2 m v_i ^2 ## = change in kinetic energy

For a system of particles,

Let's say that the work - energy theorem holds for tha j_th particle,

##W_j = ( K_f)_j - (K_i) _j##

For n - particles ,

##Σ_{ j = 1} ^n W_j = Σ_ {j = 1} ^n ( K_f)_j - Σ_{ j = 1} ^n (K_i) _j ##

So, the sum of work done on all particles = sum of change in kinetic energy of all particles

⇒ work done on a system of particles = change in kinetic energy of the system of particles

where, work done on a system of particles ≡ sum of work done on each individual particle

kinetic energy of the system of particles ≡ sum of kinetic energy of each individual particle

Now, a rigid body is nothing but a special case of a system of particles, so what is valid for a system of particles is valid for a rigid body.

Here, I feel tempted to use the same analogy for a system of particles, too.

B) Check whether it is applicable for a system of particles and a rigid body

**Work - energy theorem for one particle system,**total sum of work done by individual forces = work done by total force

To show the above equality,

let's say that there acts n forces on a particle and the net displacement of the particle is from ## \vec s_i ## to ## \vec s_f##.

Then , total sum of work done by individual forces = ## Σ_{i =1} ^n W_i = Σ_{i =1} ^n \int _ {\vec s_i } ^ {\vec s_f} \vec F_i . d\, \vec s ##

## = \int _ {\vec s_i } ^ {\vec s_f} \{Σ_{i =1} ^n \vec F_i\} . d\, \vec s ##

## \vec F_{net} = Σ_{i =1} ^n \vec F_i ##

So, total sum of work done by individual forces = ## \int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s ## = work done by total force

Now, the work energy theorem says ,

The total work done on a particle by net force is equal to change in its kinetic energy provided that the mass of the particle remains constant.

## W = \int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s = \int _ {t_i } ^ {t_f} m \frac {d \vec v } {dt} . \vec v d\, t ##

## = \int _ {v_i } ^ {v_f} m \frac 1 2 d\, v^2 = ## ## \frac 1 2 m v_f ^2 - \frac 1 2 m v_i ^2 ## = change in kinetic energy

For a system of particles,

Let's say that the work - energy theorem holds for tha j_th particle,

##W_j = ( K_f)_j - (K_i) _j##

For n - particles ,

##Σ_{ j = 1} ^n W_j = Σ_ {j = 1} ^n ( K_f)_j - Σ_{ j = 1} ^n (K_i) _j ##

So, the sum of work done on all particles = sum of change in kinetic energy of all particles

⇒ work done on a system of particles = change in kinetic energy of the system of particles

where, work done on a system of particles ≡ sum of work done on each individual particle

kinetic energy of the system of particles ≡ sum of kinetic energy of each individual particle

Now, a rigid body is nothing but a special case of a system of particles, so what is valid for a system of particles is valid for a rigid body.

Here, I feel tempted to use the same analogy for a system of particles, too.

**Can I say that what is valid for a single particle is valid for a system of particles and so, is valid for a rigid body, but not the vice - versa?****So, the answer to B is " yes".****Are the above arguments correct?**