Question

Prove that work energy theorem by calculus method.

Ans:- The time rate of change of kinetic energy is

d (K) /dt = (d(1/2) mv2)/dt)

= m(dv/dt) v

F = mv (From Newton’s second law)

= F( dx/dt)

Thus dK = Fdx

Integrating from the initial position (x i ) to final position ( x f ), we have

Ki ∫Kf dK = xi ∫xf F dx

where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f



(K f - KI) = xi ∫xf F dx

Using W = xi ∫xf F dx



(K f - KI) = W

The WE theorem is proved for a variable force.​

Answer 1

Answer:Work energy theorem states that the change in kinetic energy of an object  

is equal to the net-work done on it by the net force.

Let us suppose that a body is initially at rest and a force.  

F

 

is applied on the body to displace it through

d  

s

 

along the direction of the force. Then, a small amount of work done is given by

dw =  

F

.d  

s

 = Fds

Also, according to Newton's second law of motion, we have

F = ma  

where a is acceleration produced (in the direction of force) on applying the force. Therefore,  

dw = Mada = M  

dt

dv

​  

ds

Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given by

W = ∫  

u

v

​  

Mvdv = M∫  

u

v

​  

vdv

    = M∣  

2

v  

2

 

​  

∣  

u

v

​  

 

W =  

2

1

​  

Mv  

2

−  

2

1

​  

Mu  

2

 

Hence, work done on a body by a force is equal to the change in its kinetic energy.

Explanation:

Answer 2

Answer:

Work energy theorem => Kf-Ki =W

Explanation:

Solution:

Work done by all the forces additionally is equal to the change in kinetic energy.

W (all forces) =$\int\limits^ {} \, f.dx$

                      =$m\int\limits ^a_b {vdv/dx} *\, dx \\\\m\int\limits^v_u {vdv} \, \\m[v2/2-u2/2]\\\\1/2mv2-1/2mv2\\=kf-ki$

So, W (all forces)=ΔK (CHANGE IN THE KINETIC ENERGY)

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