Kinetic Energy

Kinetic energy is a core principle in physics that quantifies the energy possessed by an object in motion. It is essential for grasping a range of physical phenomena, from the motion of everyday items to the behavior of celestial bodies. Comprehending kinetic energy is crucial in disciplines such as mechanics, engineering, and sports science, as it provides insights into how objects move and interact in our environment.

1.0Kinetic Energy Definition

• It is the inherent ability of an object to perform work due to its motion.
• The K.E of a body can be calculated by the amount of work done in stopping the moving body or by the amount of work done in imparting the present velocity to the body from the state of rest.
• K . E$=\frac{1}{2}m{v}^2$

2.0Examples Of Kinetic Energy

1. A bullet shot from a gun can penetrate a target because of its K.E
2. The K.E of a fast stream of water is used to run water mills

3.0Expression For Kinetic Energy

$m=\text{ mass of the body, }u=\text{ o initial velocity of the body },F=\text{ constant force applied on the body }$

$a=\text{ acceleration produced in the direction of force },v=\text{ final velocity },s=\text{ distance covered }$

$v^2-u^2=2as$

$v^2-0^2=2as$

$a=\frac{v^2}{2s}$

$W=Fs=ma\cdot s=m\cdot \frac{v^2}{2s}\cdot s=\frac{1}{2}m{v}^2$

Work appeared as Kinetic Energy, K . E= $\frac{1}{2}m{v}^2$

Expression For Kinetic Energy (Calculus Method):

A body of mass m primarily at rest, a force F is applied on the body produces a displacement ds in the same direction,small work done is given by,

$dW=\vec{F}\cdot \overrightarrow{ds}=Fds\mathrm{Cos}{0}^{\circ }=Fds$

According to Newton's Second Law of Motion,

F=m a=m $\frac{dv}{dt}$

$dW=Fds=m\frac{dv}{dt}\cdot ds=mvdv$

$\int dW={\int }_0^{v}mvdv=m{\int }_0^{v}vdv=m{\left[\frac{v^2}{2}\right]}_0^v=\frac{1}{2}m{v}^2$

Work appeared as Kinetic Energy,

$K\cdot E=\frac{1}{2}m{v}^2$

4.0Work-Energy Theorem

It states that work done by all the forces (internal and external) on a particle equals to change in its kinetic energy.

$W_{\text{all }}=\Delta K\cdot E$

$W_{\text{all }}=K_f-K_i=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Proof:

1. For Constant Force:

$v^2=u^2+2as$

$v^2-u^2=2as$

$v^2-u^2=2\left(\frac{F}{m}\right)s$

$S=\frac{\left(v^2-u^2\right)m}{2F}$

$S=\frac{\Delta K\cdot E}{F}$

$Fs=\Delta K.E$

$W=\Delta K.E$

2. For Variable Force:

$W=\int Fdx=\int madx=\int m\left(\frac{vdv}{dx}\right)dx={\int }_u^{v}mvd$

$v=m{\left[\frac{v^2}{2}\right]}_u^v=\frac{m}{2}\left[v^2-u^2\right]$

$W=\Delta K\cdot E$

5.0How to apply Work – Energy Theorem

• The work-kinetic energy theorem is deduced here for a single body moving relative to an inertial frame, therefore it is recommended at present, to use it for a single body in an inertial frame. To use work-kinetic energy theorem the following steps should be followed:
•  Identify the initial and final positions as position 1 and 2 and write expressions for kinetic energies, whether known or unknown.
• Draw the free body diagram of the body at any intermediate stage between positions 1 and 2.The forces shown will help in deciding their work. Calculate work by each force and add them to obtain the total work done W12 by all the forces.
• Use the work obtained in step 2 and kinetic energies obtained in step 1 in the equation.

6.0Relation Between Kinetic Energy and Momentum

$K.E=\frac{1}{2}m{v}^2$

Momentum p=m v

Hence, $K=\frac{p^2}{2m}$

$p=\sqrt{2mk}$

$K\propto p^2\Rightarrow \frac{K_1}{K_2}=\frac{p_1^2}{p_2^2}$

For small changes(<5%);

$\frac{\Delta K}{K}=2\frac{\Delta p}{p}$

7.0Sample Questions on Kinetic Energy

Q-1.Which body has greater kinetic energy when two objects of different masses share the same linear momentum?

Solution: $K=\frac{p^2}{2m}$,for constant linear momentum $K\propto \frac{1}{m}$

Therefore, the lighter body possesses more kinetic energy than the heavier body.

Q-2. A truck and a car are traveling with the same kinetic energy on a straight road. If their engines are turned off at the same time, which vehicle will come to a stop in a shorter distance?

Solution: Applying work-energy theorem,

Loss in K.E of the vehicle=Work done against force of friction ✕ Distance

$K.E=f\times s=\mu R\times s=\mu \mathrm{mg}s$

$S=\frac{K\cdot E}{\mu mg}\Rightarrow S\propto \frac{1}{m}$

As the truck has more mass than the car ,it will stop in a lesser distance than the car.

Q-3. 0A force applied on an object of mass 2 kg changes its velocity from $(2\hat{i}+6\hat{j})$m/s to $(3\hat{i}+4\hat{j})\mathrm{m}/\mathrm{s}$. Find work done in the process?

Solution:

$v_i=\sqrt{(2{)}^2+(6{)}^2}=\sqrt{40}\mathrm{m}/\mathrm{s}$

$v_f=\sqrt{(3{)}^2+(4{)}^2=\sqrt{25}}=5\mathrm{m}/\mathrm{s}$

$W=\Delta K\cdot E=\frac{1}{2}m\left(v_f^2-v_i^2\right)=\frac{1}{2}\times 2\left(5^2-(\sqrt{40}{)}^2\right)=15\mathrm{J}$

Q-4. If momentum is increased by 300%,Find percentage change in Kinetic Energy?

Solution: Percentage change in kinetic energy,

$\frac{K\cdot E_f-K\cdot E_i}{K\cdot E_i}\times 100$

$K.E_i=\frac{p^2}{2m}$

$K\cdot E_f=\frac{(4p{)}^2}{2m}$

Momentum is increased by 300% $\left(P_f=4P_i\right)$

$\frac{\frac{p^2}{2m}[16-1]}{\frac{p^2}{2m}}\times 100=1500\%$

Percentage change in Kinetic Energy=1500 %