Work, Energy and Power Concept Page - 4

Definition

Calculation of Potential Energy

For an object having mass m$m$ having its center of gravity at height h$h$ and g$g$ be the acceleration due to gravity.
P.E.=mgh$P.E.=mgh$

Result

Relationship between force and potential energy

Force is equal to the gradient of potential energy.
F⃗ =−δUδxi^−δUδyj^−δUδzk^$\overrightarrow{F}=-{\displaystyle \frac{\delta U}{\delta x}}\hat{i}-{\displaystyle \frac{\delta U}{\delta y}}\hat{j}-{\displaystyle \frac{\delta U}{\delta z}}\hat{k}$
Example:
A particle of mass 1$1$ kg is moving in the x−y$x-y$ plane and its potential energy U$U$ in joule obeys the law U=6x+8y$U=6x+8y$, where (x,y$x,y$) are the coordinates of the particle in meter. If the particle starts from rest at (9,3) at time t=0$t=0$, then find the force on the body.Solution:
F⃗ =−δUδxi^−δUδyj^−δUδzk^$\overrightarrow{F}=-{\displaystyle \frac{\delta U}{\delta x}}\hat{i}-{\displaystyle \frac{\delta U}{\delta y}}\hat{j}-{\displaystyle \frac{\delta U}{\delta z}}\hat{k}$
F⃗ =−6i^−8j^$\overrightarrow{F}=-6\hat{i}-8\hat{j}$

Definition

Problem based on Potential Energy

Example: A bag of wheat weighs 400$400$ kg. To what height should it raised so that its potential energy is 9800$9800$ joule (g=9.8ms−2$g=9.8m{s}^{-2}$)
Solution: P.E=mgh$P.E=mgh$
             9800=400×9.8×h$9800=400\times 9.8\times h$
             h=2.5 m$h=2.5m$

Example

Work Energy theorem with multiple external forces

Question: A body is moving up on inclined plane of angle θ$\theta$ with an initial kinetic energy E$E$. The coefficient of friction between the plane and the body is μ$\mu$. What is the work done against friction before the body comes to rest?

Work energy theorem
12mu2−12mv2=W=${\displaystyle \frac{1}{2}\mathrm{m}{\mathrm{u}}^2-\frac{1}{2}\mathrm{m}{\mathrm{v}}^2=\mathrm{W}=}$ F. S$\mathrm{S}$
a=−(gsinθ+μkgcosθ)$\mathrm{a}=-(\mathrm{g}\sin \theta +{\mu }_{\mathrm{k}}\mathrm{g}\cos \theta )$
v2−u2=2as${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{a}\mathrm{s}$
s=Emg(sinθ+μkcosθ)$s={\displaystyle \frac{E}{mg(sin\theta +{\mu }_{k}cos\theta )}}$
f=μkmgcosθ$f={\mu }_{k}mgcos\theta$
W =$=$ F.S$F.S$.=μEcosθμcosθ+sinθ$={\displaystyle \frac{\mu E\cos \theta }{\mu \cos \theta +\sin \theta }}$

Example

Use of Work Energy Theorem in Pulley Mass System

Question. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg$0.36kg$ and 0.72 kg$0.72kg$. Taking g=10$g=10$ m/s2${}^2$, find the work done (in joules) by the string on the block of mass 0.36$0.36$ kg during the first second after the system is released from rest.

Solution:
2mg−T=(2m)a$2mg-T=(2m)a$
T−mg=ma$T-mg=ma$
⇒a=g/3$\Rightarrow a=g/3$
T=4mg3$T={\displaystyle \frac{4mg}{3}}$
w=T×s=T× 12at2=4mg3×12×g3×12=2009×0.36=8Joules$\mathrm{w}=\mathrm{T}\times s=\mathrm{T}\times {\displaystyle \frac{1}{2}\mathrm{a}{\mathrm{t}}^2=\frac{4mg}{3}\times \frac{1}{2}\times \frac{g}{3}\times 1^2=\frac{200}{9}\times 0.36=8Joules}$

Example

Work done by an impulsive force

Example:
A batsman hits a ball of mass m moving with an initial velocity of u. After the impact, the direction of motion of the ball reverses and velocity becomes v. Find the work done by the batsman on the ball.
Solution:
Change in kinetic energy , ΔK=12mv2−12mu2$\mathrm{\Delta }K={\displaystyle \frac{1}{2}}m{v}^2-{\displaystyle \frac{1}{2}}m{u}^2$
By work-energy theorem, work done by batsman is, W=ΔK=12mv2−12mu2$W=\mathrm{\Delta }K={\displaystyle \frac{1}{2}}m{v}^2-{\displaystyle \frac{1}{2}}m{u}^2$

Example

Change in energy of finite length objects

Question: A uniform chain is held on a frictionless table with half of its length hanging over the edge. If the chain has a mass m$m$ and length L$L$,how much work is required to pull the hanging part back onto the table?
(Given acceleration due to gravity =g$=g$)

Solution:
Work is required to pull against the gravity.Let ρ$\rho$ be linear density  ρ=mL$\rho ={\displaystyle \frac{m}{L}}$. Let x=0 at the bottom most point of the hanging chainThe displacement of each differential element of mass dm at a distance x from the bottom = L2−x${\displaystyle \frac{L}{2}}-x$
Work done W= ∫L/20Fdx=∫L/20−dm×g×(L2−x)∫L/20−ρg(L2−x)dx=−ρg(Lx2−x22)L/20=−ρg8=mgL8$$\begin{gathered} {\int }_0^{L/2}Fdx={\int }_0^{L/2}-dm\times g\times ({\displaystyle \frac{L}{2}}-x) \\ {\int }_0^{L/2}-\rho g({\displaystyle \frac{L}{2}}-x)dx \\ ={-\rho g({\displaystyle \frac{Lx}{2}}-{\displaystyle \frac{x^2}{2}})}_0^{L/2} \\ ={\displaystyle \frac{-\rho g}{8}} \\ ={\displaystyle \frac{mgL}{8}} \end{gathered}$$

Example

Application of work-energy theorem in non-inertial frame of reference

Example:
A body of mass 1 kg$1kg$ is in rest inside a vehicle on a friction less surface. When the vehicle starts moving with an acceleration of 1m/s2$1m/s^2$, the body moves a distance of 1 m$1m$ in the backward direction. Find the speed of the body at this point.
Solution:
Force on the body in the vehicle frame of reference, F=ma=1N$F=ma=1N$
Work done by this force, W=Fs=1J$W=Fs=1J$ 
This equals change in kinetic energy by work-energy theorem.
KEf−KEi=W$K{E}_f-K{E}_i=W$
12×1×v2−0=1${\displaystyle \frac{1}{2}}\times 1\times v^2-0=1$
v=2√m/s$v=\sqrt{2}m/s$

Definition

Work done by string

There are two ways to find out the work done by the string:

• By observation on inextensible strings of constant lengths: In this method, we move any one block of system observe how the length of string is being shifted from one side of pulley to other side and analyze the motion of movable pulley too. In case of a moving pulley, it pulls double the length of string from either side whereas in case of a fixed pulley change in length in string on one side is compensated by change in length of string on the other side.
• By method of Virtual Work: In this method, we consider total work done by ideal string to be zero in displacement of blocks of system, (as we know ideal string is massless which can never gain or supply energy). To analyze the same we consider different displacements of the blocks and calculate work done by strings on blocks by taking scalar product of string tension to the displacement of the block and equate total work to zero which gives us the relation of displacements of all the blocks of the system which, in turn gives us the relation in velocity and accelerations.

Definition

Internal and external forces

External forces are forces caused by external agent present outside of the system. External non-zero net force imparts an acceleration to the center of mass of the system regardless of point of application.
Internal forces are forces exchanged by the objects in the system. Internal forces may cause acceleration in different parts of the system but does not cause any acceleration in the center of mass of the entire system. 
Example:
Friction is an external force if the body experiencing friction in the system.
If both the bodies involved in friction are considered as a system, then it acts as an internal force.

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