physics-class-11-unit-06-chapter-03-work-energy-and-power-example-problems-lecture-3-6-w0whwn4lyg0

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Power </primary>
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- Power: <span style="color:green">Rate at which work is done</span>.

- $\Delta w=\vec{F} \cdot \Delta \vec{r}$

- $\frac{\Delta W}{\Delta t}=\vec{F} \cdot \frac{\Delta \vec{r}}{\Delta t}$

- Power = <span style="color:green">$\vec{F} \cdot \vec{v}$</span>

- <span style="color:blue">$\vec{v}$ instantaneous velocity.</span>

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<footer style = "font-size: 18px"> $\rightarrow$ Work, Energy and Power $\rightarrow$ <span style = "color:blue"> Power </span> $\rightarrow$ Power $\rightarrow$ Power </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Problem Solving Using Energy Methods </primary>
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- Any force which is perpendicular to the path of the particle does not contribute to the work done by external forces on the particle.

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<footer style = "font-size: 18px">Problem Solving Using Energy Methods $\rightarrow$ Problem Solving Using Energy Methods $\rightarrow$ <span style = "color:blue"> Problem Solving Using Energy Methods </span> $\rightarrow$ Work Energy Theorem $\rightarrow$ Work Energy Theorem </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Theorem </primary>
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- <span style="color:green">$\Delta K = W$ by all external forces.
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- Sum of all the work done is equal to the change in kinetic energy $\rightarrow$ <span style="color:green">$\left(K_2-K_1\right)$.</span>

- $W=-\Delta$ <span style="color:blue">(Potential energy)</span>

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<footer style = "font-size: 18px">Problem Solving Using Energy Methods $\rightarrow$ Problem Solving Using Energy Methods $\rightarrow$ <span style = "color:blue"> Work Energy Theorem </span> $\rightarrow$ Work Energy Theorem $\rightarrow$ Work Energy Theorem </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Theorem </primary>
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- $W=-\Delta(\vec{V})$

- The work done by conservative forces, only depends on state 1 and state 2 and it does not depend on the path taken between 1 and 2.

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<footer style = "font-size: 18px">Problem Solving Using Energy Methods $\rightarrow$ Work Energy Theorem $\rightarrow$ <span style = "color:blue"> Work Energy Theorem </span> $\rightarrow$ Work Energy Theorem $\rightarrow$ Work Energy Theorem </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Particle Moves in a Curve Path </primary>
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- Particle moves in a curve path.

- Position 1, the normal component of acceleration, <span style="color:green">$ a_n=\frac{v^2}{R}$</span>.

- where:

- v is the speed of the particle
 
- R is the radius of curvature.

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<footer style = "font-size: 18px">Work Energy Theorem $\rightarrow$ Work Energy Theorem $\rightarrow$ <span style = "color:blue"> Particle Moves in a Curve Path </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- There are four forces acting. <span style="color:green">Normal reaction, weight, friction and the force F</span>, which is being applied on the block, which is pushing it up on the incline.

- For gravity, which is a constant force, the acceleration is constant.

- Work done by N = 0.

- Any force which is perpendicular to the path of the particle does not contribute to the work done by external forces on the particle.

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<footer style = "font-size: 18px">Work Energy Theorem $\rightarrow$ Particle Moves in a Curve Path $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Example 1. A block of mass $5 \mathrm{kg}$ is fired up a $30^{\circ}$ incline with a speed $v_0$=5m/s. It radius $a$ distance d=2m up the incline, it comes to rest for a moment and then slides back to the bottom. Find the velocity of the block when it comes down at its starting position.

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<footer style = "font-size: 18px">Particle Moves in a Curve Path $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ External forces </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- <span style="color:green">Friction force</span> between the block and the ground.

- We will assume this friction force is not equal to <span style="color:green">0</span>.

- Draw the free body diagram of this block.

- $v_0$=5m/s, it travels the distance d = 2m, and angle is given to be = $30^{\circ}$.

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ External forces $\rightarrow$ Work Energy Method </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> External forces </primary>
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- 1) <span style="color:green">Friction</span> f $=\mu N$ <span style="color:green">(the block is sliding)</span>

- 2) m g, which is acting vertically down.

- 3) $\mathrm{N}$,  normal reaction.

- N = mg $ \cos\theta :  \Sigma F_y= 0$ 
 
-  -mg $ \sin\theta - f = ma_x$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> External forces </span> $\rightarrow$ Work Energy Method $\rightarrow$ Work Energy Method </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Method </primary>
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- Starting point : 1

- Point where v = 0 : 2
 
- $\Delta K+\Delta V=W_{\text{non-conservative forces}} $ 
 
- $\Delta K=K_2-K_1=\frac{1}{2} m(0)^2-\frac{1}{2} m v_0^2$
 
- $(\Delta v)=V_2-V_1=$ $m gd \cos\theta-0$

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<footer style = "font-size: 18px">Problem $\rightarrow$ External forces $\rightarrow$ <span style = "color:blue"> Work Energy Method </span> $\rightarrow$ Work Energy Method $\rightarrow$ Work Energy Method </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Method </primary>
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- $W_{\text {friction }}=-f d$
 
- $-\frac{1}{2} m v_0^2+m g d \sin \theta=-f d$
 
- $f=\left(\frac{m v_0^2}{2 d}-m g \sin \theta\right)$
 
- $\Delta K=K_2-K_1=\frac{1}{2} m(O)^2-\frac{1}{2} m v_0^2$
 
- $(\Delta v)=v_2-v_1=m g d \sin \theta-0$

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<footer style = "font-size: 18px">External forces $\rightarrow$ Work Energy Method $\rightarrow$ <span style = "color:blue"> Work Energy Method </span> $\rightarrow$ Work Energy Method $\rightarrow$ Work Energy Method </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Method </primary>
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- We use the work energy principle between 1 and 3.
 
- Friction force has the same magnitude. 
 
- $|F|=\mu_k N$, $N = m g \cos \theta$ 
 
- The work done by friction from 2 to 3 = - fd.

- The work done by friction during its movement from 1 to 2 = -fd
 
- When the particle moves from 1 to 3 the work done by friction,

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<footer style = "font-size: 18px">Work Energy Method $\rightarrow$ Work Energy Method $\rightarrow$ <span style = "color:blue"> Work Energy Method </span> $\rightarrow$ Work Energy Method $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Work Energy Method </primary>
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- $W_{\text {friction }} =-f d+(-f d)=-2 fd$

- $\Delta K+\Delta V=W_{\text {friction }}$

- $\Delta K = \frac{1}{2} m v_3^2-\frac{1}{2} m v_0^2$
 
- Potential energy at 3, <span style="color:green">$V_3=0$ </span>

- Potential energy at 1, <span style="color:green">$V_1=0$ </span>

- $\frac{1}{2} m\left(v_3^2-v_0^2\right)=-2 f d$
 
- $v_3=3.77 \mathrm{~m} / \mathrm{s}$  $\uparrow$

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<footer style = "font-size: 18px">Work Energy Method $\rightarrow$ Work Energy Method $\rightarrow$ <span style = "color:blue"> Work Energy Method </span> $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- A block or a particle moving in a circular path.

- Mass m tied to a string of length l.

- Assume frictionless path.

- String applies a tension force T.

- If the particle has to maintain contact, the normal reaction has to act downwards.

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<footer style = "font-size: 18px">Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ <span style = "color:blue"> Motion in a Vertical Circle </span> $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ Condition to Complete Circular Loop </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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-  a) Find the minimum velocity at the top so the particle can rotate in a full circle.

-  b) Minimum velocity at bottom for (a) to happen.
 
-  Find $v_{\text {bottom }}$, so that the particle undergoes a full circle motion.
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<footer style = "font-size: 18px">Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ <span style = "color:blue"> Motion in a Vertical Circle </span> $\rightarrow$ Condition to Complete Circular Loop $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Condition to Complete Circular Loop </primary>
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- $m g+T=\frac{m v_{\text {top }}^2}{R}=\frac{m v_{\text {top }}^2}{l}$

- $T \geqslant 0$
 
- T = $\frac{m v_{top}^2}{l}-m g$

- $\Rightarrow v_{\text {top }}^2 \geqslant$ lg

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<footer style = "font-size: 18px">Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ <span style = "color:blue"> Condition to Complete Circular Loop </span> $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- Minimum velocity at the top = $\sqrt{l g}$ or $\sqrt{R g}$

- where R is the radius of the circular loop.

- $v_{\text {bottom: }}$: use work energy principle

- 1: top
 
- 2: botom
 
- $K_2=\frac{1}{2} m v_b^2$,

- $K_1=\frac{1}{2} m v_t^2=\frac{1}{2} m l g$

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<footer style = "font-size: 18px">Motion in a Vertical Circle $\rightarrow$ Condition to Complete Circular Loop $\rightarrow$ <span style = "color:blue"> Motion in a Vertical Circle </span> $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- The work done by tension (T) or the normal reaction(N) = 0
 
- $\Delta K+\Delta V=0$
 
- $v_2=m g(2 l)$ or $m g(2 R)$
 
- $v_1=0$

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<footer style = "font-size: 18px">Condition to Complete Circular Loop $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ <span style = "color:blue"> Motion in a Vertical Circle </span> $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle </footer>

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- $\Delta K+\Delta V=0$
 
- $\Rightarrow$  $\frac{1}{2} m\left(v_b^2-v_t^2\right)-m g 2 l=0$
 
- $v_b=\sqrt{5 g l}$
 
-  a) This is not uniform circular motion.  
 
-  b) T $\rightarrow$ changing with position.

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- Speed at any location  $\theta$

- <span style="color:green"><ins>Work Energy Principle<ins></span>

- Draw a free body diagram and use Newton's second law along $\vec{r}$ direction
- to get the value of T or N at that $\theta$.

### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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- What is the speed at O, so that the mass m  just reaches A?

- $\frac{1}{2} m\left(v_O^2-v_A^2\right)+( 0 - mgl)=0$

- $v_O=\sqrt{2 g l}$

- $v_0 \leq \sqrt{2 gl}$
 
- Pendulum oscillates in a semicircle about 0.
 
-  $v_O \ge \sqrt{5 g l}$

- The pendulum undergoes a full circle motion.

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### <Success style="font-size:25px"> Work Energy And Power Example Problems L-3 </Success>
### <primary style="font-size:24px"> Motion in a Vertical Circle </primary>
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* $v_0$ lies between $\sqrt{2g l}$ and $\sqrt{5gl}$.
* The particle will encounter a position, T = O.
* At that point it will leave the circular path and then it will move like a projectile under influence of gravity.
* Once T = O, particle move like a projectile in a parabolic path

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<footer style = "font-size: 18px">Motion in a Vertical Circle $\rightarrow$ Motion in a Vertical Circle $\rightarrow$ <span style = "color:blue"> Motion in a Vertical Circle </span> $\rightarrow$ Spring is Connected to a Particle $\rightarrow$ Thank You </footer>