 # Work Energy Theorem

## (5) Work Energy Theorem

• We have already discussed about Work done on the object and kinetic energy
• The relationship between Work and kinetic energy of the object is called the Work Energy Theorem
• It states that the net work done on the system is equal to the change in Kinetic energy of the system
$W_{net} = \Delta K$
Where K is the Kinetic Energy of the body
• We shall now see the proof of the Work energy theorem
• We will prove it with constant force and variable force

### Work Energy Theorem for Constant Force

• Consider a body of mass m moving under the influence of constant force F.From newton's second law of motion
$F=ma$
Where a is the acceleration of the body
• If due to this acceleration a,velocity of the body increases from v1 to v2 during the displacement d then from equation of motion with constant acceleration we have
$v_2^2 -v_1^2=2ad$ or
$a= \frac {v_2^2 -v_1^2}{2d}$ Using this acceleration in Newton's second law of motion
we have
$F=m\frac {v_2^2 -v_1^2}{2d}$
or
$Fd=\frac {m(v_2^2 -v_1^2)}{2}$
or
$Fd=\frac {1}{2}mv_2^2 -\frac {1}{2}mv_1^2$           (7)
We know that Fd is the work done by the force F in moving body through distance d
• In equation 7 quantity
K2=mv22/2
is the final Kinetic energy of the body and
K1=mv12/2
is the initial Kinetic of the body .Thus equation 7 becomes
W=K2-K1=ΔK           (9)
• Where ΔK is the change in KE.Hence from equation (9) ,we see that work done by a force on a body is equal to the change in kinetic energy of the body

### Work Energy Theorem for Variable Force

• Lets consider a body is acted by the variable force
• Work done by the variable force is given by
$W= \int_{x_a}^{x_b} fdx$
• Now the kinetic energy at any instant will be given as
$K= \frac {1}{2}mv^2$
$\frac {dK}{dt} = \frac {d}{dt} \frac {1}{2}mv^2$
$\frac {dK}{dt} = m \frac {dv}{dt} v$
Now $F=ma= m \frac {dv}{dt}$
$\frac {dK}{dt} = f v$
$\frac {dK}{dt} = f \frac {dx}{dt}$
$dK = f dx$
Integrating,
$K_2 - K_1 =\int_{x_a}^{x_b} fdx$
or
$K_2 - K_1 = W$
$\Delta K = W$
• Hence net work done by a force on a body is equal to the change in kinetic energy of the body

## Conclusion on Work Energy Theorem

• If there are number of forces acting on a body then we can find the resultant force ,which is the vector sum of all the forces and then find the work done on the body
• Again equation (9) is a generalized result relating change in KE of the object and the net work done on it.This equation can be summarized as
Kf=Ki+W           (10)
which says that kinetic energy after net work done is equal to the KE before net work plus network done.Above statement is also known as work-kinetic energy theorem of particles
• Work energy theorem holds for both positive and negative work done.if the work done is positive then final KE increases by amount of the work and if work done is negative then final KE decreases by the amount of work done

Example-1
A block of mass m = 1 kg, moving on a horizontal surface with initial speed of 2 m/s enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding Force F on the block in this range is inversely proportional to x over this range
$F= - \frac {k}{x}$ for 0.1 < x < 2.01 m
= 0 for x < 0.1m and x > 2.01 m
where k = 0.5 J. What is the final kinetic energy and speed of the block as it crosses this patch ?
Solution
From Work Energy theorem
$K_f - K_i = \int_{x_a}^{x_b} fdx$
$K_f - \frac {1}{2} 1 \times 2^2 = \int_{.10}^{2.01} - \frac {k}{x} dx$
or
$K_f = 2 - k ln \frac {2.01}{.1}$
$K_f= .5 J$
Now
$\frac {1}{2}mv_f^2 = .5$
or
$v_f = \sqrt {\frac {2 \times .5}{1}}$
$v_f=1m/s$

Example-2
A Jeep having mass 1000 kg is moving with 30 m/s speed. Suddenly it applies the brakes and skids to rest in certain distance d. The friction force between the tyres and road is 6000 N. Find the value of d?
Solution
From Work Energy theorem
$K_f - K_i =F.d$
$0 - \frac {1}{2} mv^2 = -6000.d$
$\frac {1}{2} 1000 \times 30^2 = 6000d$
d=75 m