Work, Energy and Power Concept Page - 7

Example

Springs in series

When springs are connected in series, force is same in both the springs. Equivalent spring constant is given by:
1keq=1k1+1k2${\displaystyle \frac{1}{k_{eq}}}={\displaystyle \frac{1}{k_1}}+{\displaystyle \frac{1}{k_2}}$
Elongation and energy in the springs are divided in the ratio,
x1x2=E1E2=k2k1${\displaystyle \frac{x_1}{x_2}}={\displaystyle \frac{E_1}{E_2}}={\displaystyle \frac{k_2}{k_1}}$
Example:
In the given diagram, the maximum displacement on application of a force is A. Find the maximum speed of the mass attached after the external force is removed.
Total energy before removal of force, E=U=12keqA2$E=U={\displaystyle \frac{1}{2}}{k}_{eq}{A}^2$
At zero extension, E=K=12mv2$E=K={\displaystyle \frac{1}{2}}m{v}^2$
Using conservation of energy, v=keqA2m−−−−−√$v=\sqrt{{\displaystyle \frac{k_{eq}{A}^2}{m}}}$

Definition

Force due to single stretched massless spring

A smooth semicircular wire track of radius R$R$ is fixed in a vertical plane. One end of the massless spring of natural length 3R/4$3R/4$ is attached to the lowest point O$O$ of the wire track. A small ring of mass m$m$ which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P$P$ such that the spring makes an angle 60∘${60}^{\circ }$ with the vertical. Spring constant K=mg/R$K=mg/R$. The spring force is
Here, CP=CO=R$CP=CO=R$ and ∠CPO=∠POC$\mathrm{\angle }CPO=\mathrm{\angle }POC$
Thus , ΔCPO$\mathrm{\Delta }CPO$ is an equilateral triangle. So, OP=R$OP=R$
Extension of the spring , x=$x=$OP−$-$neutral length of spring=R−3R4=R4$=R-{\displaystyle \frac{3R}{4}}={\displaystyle \frac{R}{4}}$
Therefore spring force =Kx=(mg/R)(R/4)=mg4$=Kx=(mg/R)(R/4)={\displaystyle \frac{mg}{4}}$

Example

Springs in parallel

When springs are connected in parallel, extension is same in both the springs. Equivalent spring constant is given by:
keq=k1+k2$k_{eq}=k_1+k_2$
Force and energy in the springs are divided in the ratio,
F1F2=E1E2=k1k2${\displaystyle \frac{F_1}{F_2}}={\displaystyle \frac{E_1}{E_2}}={\displaystyle \frac{k_1}{k_2}}$
Example:
In the given diagram, the maximum displacement on application of a force is A. Find the maximum speed of the mass attached after the external force is removed.
Total energy before removal of force, E=U=12keqA2$E=U={\displaystyle \frac{1}{2}}{k}_{eq}{A}^2$
At zero extension, E=K=12mv2$E=K={\displaystyle \frac{1}{2}}m{v}^2$
Using conservation of energy, v=keqA2m−−−−−√$v=\sqrt{{\displaystyle \frac{k_{eq}{A}^2}{m}}}$

Result

Conservation of Mechanical Energy for freely falling object

The sum of Kinetic energy and Potential energy for a freely falling object is constant. The proof is as follows:
Let a body of mass m$m$ fall from rest at a height h$h$ under the influence of gravity. Then, at distance d$d$ from the top, its speed is given by v=2gd−−−√$v=\sqrt{2gd}$ using third equation of motion. 
At highest point, E1=P.E.=mgh$E_1=P.E.=mgh$
At middle point, E2=P.E.+K.E.=12mv2+mg(h−d)=12m(2gd−−−√)2+mg(h−d)=mgh$E_2=P.E.+K.E.={\displaystyle \frac{1}{2}}m{v}^2+mg(h-d)={\displaystyle \frac{1}{2}}m(\sqrt{2gd}{)}^2+mg(h-d)=mgh$
At lowest point, E3=KE=12mv2=12m(2gh)−−−−−√2=mgh$E_3=KE={\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{1}{2}}m{\sqrt{(2gh)}}^2=mgh$
⟹E1=E2=E3$⟹E_1=E_2=E_3$

Definition

Average Power

It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear. The instantaneous power is then the limiting value of the average power as the time interval t approaches zero.

Average Power = δWδt${\displaystyle \frac{\delta W}{\delta t}}$

Definition

Mathematical Expression of Power

Power is expressed as:
Power = F⃗ .v⃗ $\overrightarrow{F}.\overrightarrow{v}$

Example

Power in variable mass system

Example:
A pump having efficiency 75% lifts 800 kg water per minute from a 14 m deep well and throws at a speed of 18 ms−1${}^{-1}$. Find the power of the pump.
Solution:Power used to lift the water,P=dmdtgh=80060(10)(14)=56003W$P={\displaystyle \frac{dm}{dt}}gh={\displaystyle \frac{800}{60}}(10)(14)={\displaystyle \frac{5600}{3}}W$
Efficiency=Power UsedActual Power$={\displaystyle \frac{\text{Power Used}}{\text{Actual Power}}}$
⇒0.75=5600/3Pa$\Rightarrow 0.75={\displaystyle \frac{5600/3}{P_a}}$
⇒Pa=5600/30.75=2488.88W$\Rightarrow P_a={\displaystyle \frac{5600/3}{0.75}}=2488.88W$

Example

Using power in problems of conservation of energy

Example:
A crane is used to lift 1000 kg$1000kg$ of coal from a mine 100 m$100m$ deep. The time taken by the crane is 1$1$ hour. The efficiency of the crane is 80$80$%. If g=10ms−2$g=10m{s}^{-2}$, then find the power of the crane.Solution:
Power supplied=mght${\displaystyle \text{Power supplied}={\displaystyle \frac{mgh}{t}}}$
Power used by crane=mght×10080${\displaystyle \text{Power used by crane}={\displaystyle \frac{mgh}{t}}\times {\displaystyle \frac{100}{80}}}$
=1000×10×1003600×10080=10536×8W${\displaystyle ={\displaystyle \frac{1000\times 10\times 100}{3600}}\times {\displaystyle \frac{100}{80}}={\displaystyle \frac{{10}^5}{36\times 8}}W}$

Definition

Commercial unit of energy

SI unit of electrical energy is Joules.
Commercial unit of electrical energy is kilowatt-hour. It is defined as the energy consumed by a device of power 1 kW in 1 hr.
1 kWh=3.6×106 J

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