Work, Energy and Power Concept Page - 3

Example

Friction and Spring Force acting on a body

Question: A 2$2$ kg body is dropped from height of 1$1$ m onto a spring of spring constant 800$800$ N/m as shown in the figure. A frictional force equivalent to 0.4$0.4$ kgwt acts on the body. What will be the speed of the body just before striking the spring? (Take g=10 m/s2$g=10m/s^2$)

Solution: We know that the change of kinetic energy is equal to work done by the system.
 Change of Kinetic energy =12mv2−0=12mv2$={\displaystyle \frac{1}{2}}m{v}^2-0={\displaystyle \frac{1}{2}}m{v}^2$
Work done =(mg−f)h$=(mg-f)h$ , where f=0.4kgwt=0.4×10=4N$f=0.4kgwt=0.4\times 10=4N$, friction force.
Now, 12mv2=(mg−f)h${\displaystyle \frac{1}{2}}m{v}^2=(mg-f)h$
122v2=(20−4)1${\displaystyle \frac{1}{2}}2v^2=(20-4)1$
v2=16$v^2=16$
v=4m/s$v=4m/s$

Definition

Work done on a body by a variable external force

According to work-energy theorem, W=ΔKE$W=\mathrm{\Delta }KE$
This is very useful in finding work done by variable forces when the initial and final velocity of a body is given.
Example:
Under the action of a force, a 2$2$ kg body moves such that its position x$x$ as a function of time is given by x=t33$x={\displaystyle \frac{t^3}{3}}$, where x$x$ is in metre and t$t$ in second. Find the work done by the force in the first two seconds.Solution:
v=dxdt=t2${\displaystyle v=\frac{dx}{dt}=t^2}$
W=12mv2=12mt4${\displaystyle W=\frac{1}{2}m{v}^2=\frac{1}{2}m{t}^4}$
=12×2×(2)4=16J${\displaystyle =\frac{1}{2}\times 2\times (2{)}^4=16J}$

Example

Work Done as area under force Vs displacement graph

Example: A force acts on a body and displaces it in it's direction. The graph
shows the relation between the force and displacement. The work done by the force is:
Work done by force =∫F.ds$\int F.ds$ =$=$Area under F.s graph=Area of △$=Areaof\mathrm{△}$
=12×base×height$={\displaystyle \frac{1}{2}}\times base\times height$
=12×(14−2)×60$={\displaystyle \frac{1}{2}}\times (14-2)\times 60$
=360 J$=360J$

Definition

Kinetic energy

Kinetic Energy of an object is the energy of an object due to its motion. For an object of mass m$m$, it is given by KE=12mv2$KE={\displaystyle \frac{1}{2}}m{v}^2$. 
Note:
The above definition is made for a body under translational motion. There are other forms of kinetic energy also which will be discussed in the later chapters.

Definition

Experiment to show relation between K.E, mass and velocity

1)Take a wooden plank of three feet and place it on the wooden block. This arrangement is called the inclined plane.
2) Mark A,B,C,D as shown in the figure.Take a cylindrical tin of medium size(200 g) with a tight fitting lid.
3) Fill it tightly with sand. Put an empty rectangular plastic cube container near the bottom of the inclined plane.
4) Now release the cylinder from the point A of the inclined plane. The cylinder strikes the plastic cube container which is at rest. The rolling cylinder moves the plastic cube container for some distance.
5) Mark the place where the plastic container rests. Measure the distance between the original position and the new position of the plastic cube container.
6) Repeat the experiment by releasing the cylinder from different heights (B.C.D) and measure the distance. 
Conclusion: Every time the distance will be different ,hence we can say that K.E depend on the velocity of the cylinder when left from a different height because velocity will be different while falling from the inclined plane.And if repeat the same experiment with 500 g mass the plastic block will still move to different distance.From which we can conclude that The kinetic energy of a body also depends on its mass. Higher the mass of a body, higher is its kinetic energy.

Definition

Forms of Kinetic Energy

The various forms of kinetic energy are as follows:

1. Thermal-Motion of molecules in a material
2. Sound-Energy transfer by sound waves involve vibration of air molecules
3. Electrical-Drift of electrons
4. Mechanical-Motion of an object

Definition

Calculation of Kinetic Energy

Example: Calculate the kinetic energy and potential energy of the ball half way up, when a ball of mass 0.1 kg$0.1kg$ is thrown vertically upwards with an initial speed of 20ms−1$20m{s}^{-1}$.
Solution:
Total energy at the time of projection 
=12mv2=12×0.1(20)2=20J${\displaystyle =\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.1(20{)}^2=20J}$
Half way up, P.E. becomes half the P.E. at the top i.e. 
P.E.=202=10J${\displaystyle P.E.=\frac{20}{2}=10J}$ ∴K.E.=20−10=10J$\therefore K.E.=20-10=10J$.

Definition

Kinetic Energy and Momentum

K.E.=p2/2m$K.E.=p^2/2m$
K.E.:Kinetic energy,p:momentum,m:mass$K.E.:\text{Kinetic energy},p:\text{momentum},m:\text{mass}$

Example

Kinetic Energy of a System of Particles

The mass of a simple pendulum bob is 100 g$100g$.The length of the pendulum is 1 m$1m$. The bob is drawn aside from the equilibrium position so that the string makes an angle of 600${60}^0$ with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

K.EA=Δ$K.E_A=\mathrm{\Delta }$ P.E.$P.E.$  b/w  A & B
=mg(L−Lcos600)$=mg(L-L\cos {60}^0)$
=10−1×9.8×(1−12)$={10}^{-1}\times 9.8\times (1-{\displaystyle \frac{1}{2}})$
=0.49J$=0.49J$

Formula

Problem on K.E

Example:If two bodies of equal masses move with uniform velocity of v$v$ and 2v$2v$, what will be the ratio of their kinetic energies?
Solution:
K.E1=12mv2$K.E_1={\displaystyle \frac{1}{2}}m{v}^2$
K.E2=12m×4v2$K.E_2={\displaystyle \frac{1}{2}}m\times {4v}^2$ 
Ratio of Kinetic energies=K.E1K.E2$={\displaystyle \frac{K.E_1}{K.E_2}}$
                                   =14$={\displaystyle \frac{1}{4}}$

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