physics-class-11-unit-05-chapter-07-problem-solving-law-of-motion-lecture-7-8-bjkm-vxzms

### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- **<span style="color:green">Question:</span>** A block of mass $2 \mathrm{Kg}$ placed on a long frictionless horizontal table is pulled horizontally by a constant force $\mathbf{F}$. It is found to move $10 \mathrm{~m}$ in the first two seconds. Find the magnitude of the force.

- <span style="color:green">Answer:</span> $ \text { Mass, } m=2 \mathrm{kg}$,$ \text { distance } s=10 \mathrm{~m}$,$ \text { initial velocity } u=0 $

- Time taken $t=2 \mathrm{sec}$

- Then using, $s=u t+1 / 2 a t^2 $

- u = 0, $s=1 / 2 a t^2$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- Putting the values of $s$ and $t$

- $a=5 \mathrm{~m} / \sec ^2$

- Then using, $a  =F / m \text { or } F=m a $

- $F  =2 \times 5  =10 \mathrm{~N}$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- **<span style="color:green">Question:</span>** A ship of mass $2 \times 10^7$ Kg initially at rest is pulled by a force of $7 \times 10^5$ N through a distance of  3 m. Assuming that the resistance due to water is negligible, calculate the speed of the ship.

- <span style="color:green">Answer:</span>$ v=u+a t$

* $u = 0\Rightarrow v=a t \quad-(1) $
* $ a=F/ m-(2)$

- **From (1) and (2)**

* $ v=\left(\frac{F}{m}\right) \times t \quad-(3) $
* $ s=u t+1 / 2 a t^2 $

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- **<span style="color:green">Question:</span>** A block of mass $m$ slides along a floor while a force of magnitude $\mathbf{F}$ is applied to it at an angle as shown in figure. The coefficient of kinetic friction is $\mu$. Then, what will be the value of acceleration of the block.

- **<span style="color:green">Answer:</span>** $N=m g-F \sin \theta \text {-(1) }$

- For moving body.

- $F=\cos \theta-f=m a \text {-(2) }$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- Where $f=$ frictional farce

- $f=\mu N=\mu(m g-F \sin \theta)\text {-(3) }$

- With (2) and (3)

- $m a=F \cos \theta-\mu(m g-F \sin \theta)\text {-(4) }$

- Or, $a=F / m \cos \theta-\mu(g-F / m \sin \theta)\text {-(5) }$

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- **<span style="color:green">Question:</span>** A block of mass $4 \mathrm{Kg}$ rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The co-efficient of static friction between the block and the plane is 0.7 . Calculate the frictional force acting on the block.

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- <span style="color:green">Answer:</span>

- **For block at rest,**

- $ F=m g \sin \theta $

- $ F=\text { frictional force }$

- Frictional force $=m g \sin \theta$

- $ =4 \times 9.8 \times \left(\frac{1}{2}\right) $

- $ =19.6 \mathrm{M}$

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- **<span style="color:green">Question:</span>** The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, what should be the value of angle $\theta$.

- <span style="colorgreen">Answer</span> For system in equilibrium,

- $ T_1=m g  ; \quad T_2=m g -(1)$

- $ T_1 \cos \theta_1+T_2 \cos \theta=\sqrt{2} m g -(2) $

- $ T_1 \sin \theta_1=T_2 \sin \theta \quad \text { -(3) }$

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- From (1) and (3)

- $m g \sin \theta_1=m g \sin \theta $

- $\text { or } \quad \theta_1=\theta -(4)$

- Using (4) and (1). (2) can be written as, $m g \cos \theta+m g \cos \theta=\sqrt{2} m g$

- or $2 \cos \theta=\sqrt{2} $

- $\theta=45^{\circ}$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- **<span style="color:green">Question:</span>** What is the maximum value of the force F such that the block shown in the arrangement, does not move?

- **<span style="color:green">Answer:</span>**

- $N=m g+F \sin 60^{\circ} -(1)$

- Frictional force $=F \cos 60^{\circ}  -(2)$

- Frictional force $=\mu N$

- $=\mu\left[m g+F \sin 60^{\circ}\right] -(3)$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- From (3) and (1)

- $\mu\left[m g+F \sin 60^{\circ}=F \cos 60^{\circ}\right].$

- Or $ F  =\frac{\mu m g}{\cos 60^{\circ}-\mu \sin 60^{\circ}}$

- $=\frac{(1 / 2 \sqrt{3}) \sqrt{3} \times 9.8}{(1 / 2)-\{(1 / 2 \sqrt{3}) \times(\sqrt{3} / 2)\}}$

- $=\frac{4.9}{\left(1/2-1 / 4\right)}$ = 19.6 N

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- **<span style="color:green">Question:</span>** Two particles of mass $\mathbf{m}$ each are tied at the end of a light string of length $\mathbf{2 a}$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the center P. Now the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. What will be the value of acceleration, when the separation between them is $\mathbf{2 x}$ ?

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- Net force at. $P$ must be equal to zero for this,

- $F=R=2 T \sin \theta-(1)$

- For mass $m$, with no vertical movement

- $T \sin \theta=m g  -(2)$

- $T \cos \theta=m a  -(3)$

- from (1) and (3)

- $m a  =\frac{F}{2} \cot \theta $

- $a=\frac{F}{2 m}\left(\frac{x}{\sqrt{q^2-x^2}}\right)$

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- **<span style="color:green">Question:</span>** A block of mass $2 \mathrm{Kg}$ slides on an inclined plane which makes an angle $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the surface is $\sqrt{3} / \sqrt{2}$. What force should be applied to the block so that it moves,
- (a) down and (b) up without any acceleration

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- <span style="color:green">Answer:</span>

- (b) Moving up.

- $F  =m g \sin 30^{\circ}+f $

- $ =m g \sin 3{ }^{\circ}+\mu m g \cos 30^{\circ} $

- $ =m g\left(\sin 30^{\circ}+\mu \cos 30^{\circ}\right)$

- Putting the values of $m, g$, and $\mu$

- $F=2 \times 9.8\left[(1 / 2)+(\sqrt{3} / \sqrt{2})\left(\frac{\sqrt{3}}{2}\right)\right]$

- = 30.58 N

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- **<span style="color:green">Question:</span>** In the figure, the blocks A, B and C have masses 3kg, 4 kg and 8 kg respectively. The coefficient of sliding friction between any two surfaces is 1.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected bya alight flexible cord passing around a fixed frictionless pulley, Find the force F necessary to drag C along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the figure, i.e, B on C and A on B, is maintained throughout

- Masses of the bodies

- $ A=m_A=3 \mathrm{kg} $, $ B=m_B=4 \mathrm{kg} $, and $ C=m_C=8 \mathrm{kg} $

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- <span style="color:green">Answer :</span>

- If we pull C with Force

- F then foces will act on B and C will be like this,

- T=$f_{AB} + f_{BC} -(4)$

- F =$T + f_{BC} + f_{CS} -(5)$

- F =$f_{AB} + 2 f_{BC} + f_{CS} -(6)$

- Putting the values of $f_{AB}, f_{BC}, f_{CS}$

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- <span style="color:green">Question :</span> An insect crawls up a hemispherical surface very slowly (see fig.) The coefficient of friction between the insect and the surface is 1/3. If the line joining the center of the hemispherical surface to the insect makes an angle alpha with the vertical. what will be the maximum possible value of alpha ?

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### <Success style="font-size:25px"> Problem Solving Law Of Motion L-7 </Success>
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- <span style="color:green">Solution:</span> For equilibrium,

- N = mg cos$\alpha$ -(1)

- And frictional force, f = $\mu N$ -(2)

- Also, f = mg sin$\alpha$ -(3)

- **From (1),(2) and (3)**

* $\mu (mg \cos\alpha) = mg \sin\alpha$
* $\mu = tan\alpha$
* For $\mu = \frac{1}{3}$
* tan$\alpha=\frac{1}{3}$ or $cot \alpha = 3$

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