- #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 bodyWork - 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 forceNow, 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 energyFor 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?
B) Check whether it is applicable for a system of particles and a rigid bodyWork - 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 forceNow, 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 energyFor 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?