Laws of Motion
In the given figure the blocks $A, B$ and $C$ weigh $4 \mathrm{~kg}, 6 \mathrm{~kg}$ and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5 . The force $\vec{F}$ required to slide the block $C$ with constant speed is $\_\_\_\_$ N . (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )
Explanation:
If Block C moves to the Left with velocity v. Then string pulls Block B to the Right with velocity v.
Since Block A is just sitting on Block B and the motion of block C is uniform motion so there will be kinetic friction between A and B .
For block $\mathrm{A}, \mathrm{N}_{\mathrm{AB}}-\mathrm{m}_{\mathrm{A}} \mathrm{g}=0 \Rightarrow \mathrm{~N}_{\mathrm{AB}}=\mathrm{m}_{\mathrm{A}} g$
So, the normal force between blocks A and B is $\mathrm{m}_{\mathrm{A}} \mathrm{g}=4 \times 10 \mathrm{~N}=40 \mathrm{~N}$
So, the friction between A and B is, $f_{A B}=\mu N_{A B}=0.5 \times 40 N=20 N \Rightarrow f_{A B}=20 N$
For block B,
$ -\mathrm{N}_{\mathrm{BA}}+\mathrm{N}_{\mathrm{BC}}-\mathrm{m}_{\mathrm{B}} \mathrm{~g}=0 $
$\Rightarrow-\mathrm{m}_{\mathrm{A}} \mathrm{~g}+\mathrm{N}_{\mathrm{BC}}-\mathrm{m}_{\mathrm{B}} \mathrm{~g}=0 $
$\Rightarrow $ $ \mathrm{N}_{\mathrm{BC}}=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{g} $
So, the normal force between blocks B and C is $\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{g}=(4+6) \times 10 \mathrm{~N}=100 \mathrm{~N}$
Block C moves left and B moves right. So, on block B friction acts to the left (opposing its rightward motion). And on block C friction acts to the right (opposing its leftward motion relative to B ).
$ \mathrm{f}_{\mathrm{BC}}=0.5 \times 100=50 \mathrm{~N} \Rightarrow \mathrm{f}_{B C}=50 \mathrm{~N} $
Ground supports all the three blocks, so the normal force on the ground is -
$ \mathrm{N}_{\mathrm{C}}=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{C}}\right) \mathrm{g} $
$\Rightarrow $$ N_C=(4+6+8) \times 10=18 \times 10=180 N $
Friction at ground ( $\mathrm{f}_{\mathrm{g}}$ ) acts on Block C to the Right (opposing motion).
$ \mathrm{f}_{\mathrm{g}}=0.5 \times 180=90 \mathrm{~N} \Rightarrow \mathrm{f}_{\mathrm{g}}=90 \mathrm{~N} $
As block B is moving at constant speed, so the net force is zero.
$\mathrm{T}=\mathrm{f}_{A B}+\mathrm{f}_{\mathrm{BC}}$
$\Rightarrow $ $ \mathrm{T}=20 \mathrm{~N}+50 \mathrm{~N}=70 \mathrm{~N} $
Also block C is moving at constant speed, so the net force is zero.
$ \mathrm{F}=\mathrm{T}+\mathrm{f}_{\mathrm{BC}}+\mathrm{f}_{\mathrm{g}} $
$\Rightarrow $ $\mathrm{F}=70 \mathrm{~N}+50 \mathrm{~N}+90 \mathrm{~N}$
$\Rightarrow $ $ \mathrm{F}=210 \mathrm{~N} $
So, the force required is 210 N.
A block takes $t$ time to slide down a plane inclined at $45^{\circ}$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\frac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left(\frac{\alpha}{100}\right)$. The value of $\alpha$ is $\_\_\_\_$ .
Explanation:
When a block of mass $m$ is placed on a plane inclined at an angle $\theta$ :
Case 1: Rough Surface
The forces acting along the plane are the component of gravity ( $\mathrm{mg} \sin \theta$ ) acting downwards and the kinetic friction ( $\mathrm{f}_{\mathrm{k}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta$ ) acting upwards.
The net force acting along the inclined surface is $\mathrm{F}_{\text {net }}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta$
Using Newton's second law of motion $\mathrm{F}=\mathrm{ma}$, the acceleration of block on rough surface is,
$ \mathrm{a}_{\text {rough }}=\mathrm{g}(\sin \theta-\mu \cos \theta) $
Case 2: Smooth Surface
Friction is zero ( $\mu=0$ ).
The net force acting along the inclined surface is $\mathrm{F}_{\text {net }}=\mathrm{mg} \sin \theta$
So, the acceleration of block on smooth surface is,
$ \mathrm{a}_{\text {smooth }}=\frac{\mathrm{mg} \sin \theta}{\mathrm{~m}}=\mathrm{g} \sin \theta $
The block starts from rest ( $\mathrm{u}=0$ ) and slides a distance s .
Using the kinematic equation $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2:$
$ s=\frac{1}{2} a t^2 \Rightarrow t=\sqrt{\frac{2 s}{a}} $
For a constant distance s , the relationship is $\mathrm{t} \propto \frac{1}{\sqrt{\mathrm{a}}}$, or a $\propto \frac{1}{\mathrm{t}^2}$.
The angle of inclination is $\theta=45^{\circ}$.
The time taken on rough surface is $\mathrm{t}_{\text {rough }}=\mathrm{t}$.
The time taken on smooth surface is $t_{\text {smooth }}=\frac{t}{2}$.
So, the ratio of accelerations is;
$ \frac{\mathrm{a}_{\text {smooth }}}{\mathrm{a}_{\text {rough }}}=\left(\frac{\mathrm{t}_{\text {rough }}}{\mathrm{t}_{\text {smooth }}}\right)^2=\left(\frac{\mathrm{t}}{\mathrm{t} / 2}\right)^2=(2)^2=4 $
Substituting the acceleration formulas into the ratio :
$ \frac{\mathrm{g} \sin 45^{\circ}}{\mathrm{g}\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)}=4 $
$\Rightarrow $ $\frac{1 / \sqrt{2}}{\left(\frac{1}{\sqrt{2}}-\mu\left(\frac{1}{\sqrt{2}}\right)\right)}=4$
$\Rightarrow $ $\frac{1}{1-\mu}=4$
$\Rightarrow $ $1=4-4 \mu$
$\Rightarrow $ $4 \mu=3 \Rightarrow \mu=0.75=\frac{75}{100}$
$\Rightarrow $ $\frac{75}{100}=\frac{\alpha}{100} \Rightarrow \alpha=75$
Therefore, the value of $\alpha$ is 75 .
Explanation:
$ \vec{F}_{n e t}=m \frac{d \vec{v}}{d t} $
$ \mathrm{m} \overrightarrow{\mathrm{~g}}+\overrightarrow{\mathrm{F}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} $
$ \mathrm{m} \overrightarrow{\mathrm{~g}}-\mathrm{C} \overrightarrow{\mathrm{v}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} $
Horizontal direction
$ -\mathrm{Cv}_{\mathrm{x}}=\frac{\mathrm{mdv}_{\mathrm{x}}}{\mathrm{dt}} $
$\begin{aligned} & -\frac{C}{m} \int_0^t d t=\int_{v_{0 x}}^{v_x} \frac{d v_x}{v_x} \\ & -\frac{t}{2}=\ln \frac{v_x}{v_{0 x}} \\ & \frac{d x}{d t}=v_x=v_{0 x} e^{-t / 2} \\ & \int_0^{s_x} d x=v_{0 x} \int_0^t e^{-t / 2} d t \\ & S_x=2 v_{0 x}\left(1-e^{-t / 2}\right)\end{aligned}$
At t = 2 sec
$\begin{aligned} & S_x=2 \times 270 \times \cos 60^{\circ}\left[1-\frac{1}{\mathrm{e}}\right] \\\\ & S_x=270\left(1-\frac{1}{2.7}\right) \\\\ & =\frac{270}{2.7} \times(1.7) \\\\ & =170 \mathrm{~m} \\\\ & S_x=170 \mathrm{~m}\end{aligned}$
Three blocks $\mathrm{M_1, M_2, M_3}$ having masses $4 \mathrm{~kg}, 6 \mathrm{~kg}$ and $10 \mathrm{~kg}$ respectively are hanging from a smooth pully using rope 1, 2 and 3 as shown in figure. The tension in the rope $\mathrm{1, T_1}$ when they are moving upward with acceleration of $2 \mathrm{~ms}^{-2}$ is __________ $\mathrm{N}$ (if $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$).
Explanation:
$\begin{aligned} & T_1-(4+6+10) g=(4+6+10)(2) \\ & \Rightarrow T_1=20(10+2) \\ & \quad=240 \mathrm{~N} \end{aligned}$
Two forces $\overline{\mathrm{F}}_1$ and $\overline{\mathrm{F}}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos ^{-1}\left(\frac{1}{n}\right)$. The value of $|n|$ is _______.
Explanation:
Let's denote the magnitude of the smaller force as $F$, hence the magnitude of the larger force is $3F$. The resultant force $\vec{R}$ is equal in magnitude to the larger force, which means $|\vec{R}| = 3F$. When two forces $\vec{F}_1$ and $\vec{F}_2$ act on a body, the magnitude of their resultant $\vec{R}$ can be found using the law of vector addition:
$|\vec{R}| = \sqrt{|\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2|\cos\theta}$,
where $\theta$ is the angle between $\vec{F}_1$ and $\vec{F}_2$. Given that in our case $|\vec{R}| = 3F$, $|\vec{F}_1| = F$ and $|\vec{F}_2| = 3F$, by substituting these values into the equation, we get:
$3F = \sqrt{F^2 + (3F)^2 + 2(F)(3F)\cos\theta}$
$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta$
Simplifying this equation by subtracting $10F^2$ from both sides gives:
$-F^2 = 6F^2\cos\theta$
Dividing both sides by $-F^2$ gives:
$-1 = -6\cos\theta$
Therefore, $\cos\theta = \frac{1}{6}$.
It is given that the angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$, hence comparing this with the above result, we find that $n = 6$. Therefore, $|n| = 6$.
Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t(\sqrt{2}+1)$ s. The value of t is ___________.
(use $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )
Explanation:
$AB = 10\sqrt 2 $ m
${v_A} = \sqrt {2 \times 10 \times 10} = 10\sqrt 2 $ m/s
${v_C} = 10\sqrt 2 $ m/s
${a_{BC}} = g\sin (30^\circ ) = 5$ m/s2
${t_{BC}} = 2\sqrt 2 \,s\,\left( {{{{v_c}} \over {{a_{BC}}}}} \right)$
${t_{AB}} = {{{v_A}} \over {5\sqrt 2 }} = 2\,s$
${t_{AB}} + {t_{BC}} = 2(\sqrt 2 + 1)$
$ \Rightarrow t = 2$
Four forces are acting at a point $\mathrm{P}$ in equilibrium as shown in figure. The ratio of force $\mathrm{F}_{1}$ to $\mathrm{F}_{2}$ is $1: x$ where $x=$ _____________.
Explanation:
${F_1} = \, + 2 \times {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}$
${F_2} = 2 \times {1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$
${{{F_1}} \over {{F_2}}} = {1 \over 3} = {1 \over x} \Rightarrow x = 3$
A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is ${x \over 5}$ Mg for x = ______________. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g = acceleration due to gravity)
Explanation:
$a = {{Mg} \over {4M + M}} = {g \over 5}$ (in upward direction)
$T = M\left( {g + {g \over 5}} \right) = {{6Mg} \over 5}$
$ \Rightarrow x = 6$
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta$ = tan$-$1 (x $\times$ 10$-$1). The value of x is ____________.
(Given, g = 10 m/s2)
Explanation:
The vertical component of the tension, $T\cos\theta$, balances the weight of the mass:
$T\cos \theta = mg$
Since $ mg = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$:
$T\cos \theta = 100\, \text{N}$ ...... (i)
The horizontal component of the tension is given by:
$T\sin \theta = 30\, \text{N}$ ........ (ii)
Dividing equation (ii) by equation (i):
$ \frac{T\sin \theta}{T\cos \theta} = \frac{30}{100}$
Therefore:
$ \tan \theta = \frac{3}{10}$
Hence, the value of x is:
$ \therefore x = 3$
A system to 10 balls each of mass 2 kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the 7th and 8th ball is __________ N when 6th ball just leaves the table.
Explanation:
At given instant
${a_{sys}} = {{6\,m \times g} \over {10\,m}} = {{6\,g} \over {10}}$
$\therefore$ ${T_{78}} = (3m) \times {a_{sys}}$
$ = (3m) \times \left( {{{6g} \over {10}}} \right)$
$ = {{3 \times 2 \times 6 \times 10} \over {10}} = 36\,N$
A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F = $\sqrt{x}$N as shown in figure.
The value of x = _____________.
Explanation:
$F \times {1 \over 2} = 0.2 \times 10 \times {{\sqrt 3 } \over 2}$
$ \Rightarrow F = 2\sqrt 3 $
$ \Rightarrow F = \sqrt {12} $ N
$\therefore$ $x = 12$
A force on an object of mass 100 g is $\left( {10\widehat i + 5\widehat j} \right)$ N. The position of that object at t = 2 s is $\left( {a\widehat i + b\widehat j} \right)$ m after starting from rest. The value of ${a \over b}$ will be ___________.
Explanation:
$\overrightarrow F = m\overrightarrow a $
$ \Rightarrow \overrightarrow a = 100\widehat i + 50\widehat j$
So, $\overrightarrow S = {1 \over 2}\overrightarrow a {t^2}$
${1 \over 2}\left( {100\widehat i + 50\widehat j} \right){2^2}$
$ = 200\widehat i + 100\widehat j$ m
so a = 200 m and b = 100 m
so ${a \over b} = 2$
Explanation:
On smooth incline
a = g sin30$^\circ$
by S = ut + ${1 \over 2}$at2
S = ${1 \over 2}$${g \over 2}$T2 = ${g \over 4}$T2 ........ (i)
For rough surface,
$ma = mg\sin 30^\circ - \mu mg\cos 30^\circ $
$a = g\sin 30^\circ - \mu g\cos 30^\circ $
Distance covered by the block on the rough surface in time $\alpha$T,
$s = ut + {1 \over 2}a{t^2}$
$s = 0 + {1 \over 2}(g\sin 30^\circ - \mu g\sin 30^\circ ){t^2}$
$s = {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2}$ .... (ii)
Distance covered by the block is same for both the case,
$ \Rightarrow {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2} = {g \over 4}{T^2}$ [from Eq. (i) and Eq. (ii)]
$ \Rightarrow 1 - \sqrt 3 \mu = {1 \over {{\alpha ^2}}} \Rightarrow \mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt 3 }}$
Comparing with $\mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt x }}$
The value of the x = 3.
Explanation:
Given,
Angle of inclination, $\theta$ = 30$^\circ$
Acceleration, a = 10 ms$-$2
Acceleration due to gravity, g = 10 ms$-$2
According to the question the car and bob is as shown below,
Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.
In equilibrium, $\Sigma$Fx = 0 and $\Sigma$Fy = 0
$\Rightarrow$ F' cos30$^\circ$ = T sin$\alpha$
${{ma\cos 30^\circ } \over {\sin \alpha }} = T$ ...... (i)
where, m is the mass of the bob.
$F'\sin 30^\circ + mg = T\cos \alpha $
$ \Rightarrow ma\sin 30^\circ + mg = {{ma\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$ [$\because$ using Eq. (i)]
$ \Rightarrow a\sin 30^\circ + g = {{a\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$
$10 \times {1 \over 2} + 10 = {{10 \times \sqrt 3 } \over 2}\cot \alpha $
$ \Rightarrow {1 \over 2} + 1 = {{\sqrt 3 } \over 2}\cot \alpha \Rightarrow {3 \over 2} = {{\sqrt 3 } \over 2}\cot \alpha $
or, $\cot \alpha = \sqrt 3 $
$ \Rightarrow \alpha = 30^\circ $
Explanation:
fsmax = 1a (for 1 kg block) ............ (ii)
$\mu$ $\times$ 1 $\times$ g = a
F = 15N
Explanation:
$\sqrt {{{2s} \over {{a_a}}}} = {1 \over 2}\sqrt {{{2s} \over {{a_d}}}} $ ..... (i)
${a_a} = g\sin \theta + \mu g\cos \theta $
$ = {g \over 2} + {{\sqrt 3 } \over 2}\mu g$
${a_d} = g\sin \theta - \mu g\cos \theta $
$ = {g \over 2} - {{\sqrt 3 } \over 2}\mu g$
using the above values of aa and ad and putting in equation (i) we will gate $\mu = {{\sqrt 3 } \over 5}$
Explanation:
Mbullet = 0.1 kg
V2 = 44 + 2($-$a) $\times$ 5
$ \Rightarrow $ 0 = (10)2 $-$ 2a $\times$ (0.5)
Retardation $(a) = {{1000} \over {2 \times 5}}$ = 100 m/s2
Retarding force (F) = ma = 0.1 $\times$ 100
FR = 10 N
Explanation:
$ \because $ f = T
$ \Rightarrow $ $\mu$N = T
$ \Rightarrow $ $\mu$(90 $-$ T) = T
$ \Rightarrow $ 0.5 (90 $-$ T) = T
$ \Rightarrow $ 90 $-$ T = 2T
$ \Rightarrow $ 3T = 90
$ \Rightarrow $ T = 30 N
Explanation:
Minimum possible force $ \Rightarrow $
$F = {{\mu mg} \over {\sqrt {1 + {\mu ^2}} }}$
${F_{\min }} = {{{1 \over {\sqrt 3 }} \times 1 \times 10} \over {\sqrt {1 + {1 \over 3}} }}$
Fmin = 5N
Explanation:
$a = {F \over {M + m}}$
f = m$a$ = m${F \over {M + m}}$
m${F \over {M + m}}$ $ \le $ $\mu $mg (for no slipping)
$ \Rightarrow $ F $ \le $ $\mu $(m + M)g
$ \therefore $ Fmax = ${3 \over 7}\left( {0.5 + 4.5} \right) \times 9.8$ = 21 N
Explanation:
time = 4 sec
As body start from rest therefore
position vector initially $\overrightarrow {{r_i}} = (0\widehat i + 0\widehat j + 0\widehat k)$ &
u (initial velocity) = 0
given, ${r_f} = (x\widehat i + y\widehat j + z\widehat k)$
Now, from second equation of motion
$\overrightarrow s = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$
${r_f} - {r_i} = {1 \over 2} \times \left( {{{2\widehat i + 3\widehat j + 5\widehat k} \over 2}} \right) \times {(4)^2}$
$ \Rightarrow $ $(x\widehat i + y\widehat j + z\widehat k) - (0\widehat i + 0\widehat j + 0\widehat k) = 8\widehat i + 12\widehat j + 20\widehat k$
$ \Rightarrow $ $x\widehat i + y\widehat j + z\widehat k = 8\widehat i + 12\widehat j + 20\widehat k$
$ \therefore $ The value of b = 12
Explanation:
The apparent weight $W_{\text{app}}$ of a person in an elevator moving with acceleration is given by:
$W_{\text{app}} = m(g - a)$
where:
- $W_{\text{app}}$ is the apparent weight,
- $m$ is the mass of the person,
- $g$ is the acceleration due to gravity,
- $a$ is the acceleration of the elevator.
Given that the person's mass is 60 kg, the acceleration due to gravity is 10 m/s², and the acceleration of the lift is 1.8 m/s², we can substitute these values into the formula:
$W_{\text{app}} = 60 \, \text{kg} \times (10 \, \text{m/s}^2 - 1.8 \, \text{m/s}^2) = 60 \, \text{kg} \times 8.2 \, \text{m/s}^2 = 492 \, \text{N}$
So, the apparent weight of the person when the lift descends with a uniform downward acceleration of 1.8 m/s² will be 492 N.
Explanation:
$\overrightarrow a = {{\overrightarrow F } \over m} = {{20\widehat i + 10\widehat j} \over 2} = 10\widehat i + 5\widehat j$
$ \therefore $ $\overrightarrow s = {1 \over 2}\overrightarrow a {t^2} = {1 \over 2}\left( {10\widehat i + 5\widehat j} \right) \times {\left( {10} \right)^2}$
$ = 50\left( {10\widehat i + 5\widehat j} \right)m$
$ \therefore $ Displacement along x-axis
= 50 $\times$ 10 = 500 m
[g = 10 m/s2; sin60$^\circ$ = ${{\sqrt 3 } \over 2}$; cos60$^\circ$ = ${1 \over 2}$]
Explanation:
N = Mg + Fsin60$^\circ$
N = $\sqrt 3 g + {{F\sqrt 3 } \over 2}$
For No slipping
Fcos60$^\circ$ = Friction
${F \over 2} = \mu N = {1 \over {3\sqrt 3 }}\left( {\sqrt 3 g + {{F\sqrt 3 } \over 2}} \right)$
${F \over 2} = {g \over 3} + {F \over 6}$
${F \over 2} - {F \over 6} = {g \over 3}$
${{6F - 2F} \over {12}} = {g \over 3}$
4F = 4g
F = 10
F = 3x
x = ${F \over 3}$ = ${10 \over 3}$ = 3.33
x = 3.33
Explanation:
At maximum height, the slope of tangent drawn,
$\tan \theta = {{dy} \over {dx}} = {{2x} \over 4} = {x \over 2}$ [$\because$ $y = {{{x^2}} \over 4}$]
$ \Rightarrow 0.5 = {x \over 2}$ ($\because$ $\mu$ = tan$\theta$)
$\Rightarrow$ x = 1 m
$\therefore$ $y = {{{x^2}} \over 4} = {1 \over 4}$ = 0.25 m = 25 cm
Explanation:
Various forces acting on block are shown below
Frictional force $\le$ mg
$\Rightarrow$ N $\times$ 0.2 $\le$ 5
$\Rightarrow$ N $\le$ 25
$\therefore$ Magnitude of horizontal force, F = N = 25 N
The value of t is _____________.
Explanation:
Range, $R = {u_x} \times T = {{2{u_x}{u_y}} \over g} = {{2 \times 5 \times 5} \over {10}} = 5$ m
Time of flight, $T = {{2{u_y}} \over g} = {{2 \times 5} \over {10}} = 1$ s
Both particle have no vertical velocity after splitting so both will take same time to reach the ground.
$\because$ Time of motion of one part falling vertically downwards is 0.5 s
$\Rightarrow$ Time of motion of another part, $t = 0.5$ s
The value of x is _______________.
Explanation:
Range, $R = {u_x} \times T = {{2{u_x}{u_y}} \over g} = {{2 \times 5 \times 5} \over {10}} = 5$ m
Time of flight, $T = {{2{u_y}} \over g} = {{2 \times 5} \over {10}} = 1$ s
Both particle have no vertical velocity after splitting so both will take same time to reach the ground.
$\because$ Time of motion of one part falling vertically downwards is 0.5 s
$\Rightarrow$ Time of motion of another part, $t = 0.5$ s
From momentum conservation, pi = pf
2m $\times$ 5 = m $\times$ v
v = 10 m/s
Displacement of other part in 0.5 s in horizontal direction,
$ = v\left( {{T \over 2}} \right) = 10 \times 0.5 = 5$ m = R
$\therefore$ Total distance of second part from point O is $x = {{3R} \over 2} = 3 \times {5 \over 2}$
$ \Rightarrow $ x = 7.5 m
Explanation:
Here, $\mu $s = 0.40, $\mu $k = 0.32
$ \because $ $\mu $kN3 = $\mu $sN4
and x1N3 = 40 $ \times $ N4
So, ${{{\mu _k}} \over {{x_1}}} = {{{\mu _s}} \over {40}} \Rightarrow {{0.32} \over {{x_1}}} = {{0.4} \over {40}}$
$ \Rightarrow {{0.32} \over {{x_1}}} = {1 \over {100}} \Rightarrow {x_1} = 32$ cm
${{{\mu _s}{N_5}} \over {{x_1}{N_5}}} = {{{\mu _s}{N_6}} \over {{x_g}{N_6}}} \Rightarrow {{0.40} \over {10 \times 32}} \times {{0.32} \over {{x_g}}}$
$ \Rightarrow {x_g} = 25.6$
Explanation:
The block was initially at rest and its velocity just after the application of impulse is $v(0) = {v_0}{e^{ - 0/\tau }} = {v_0}$. The applied impulse is equal to the change in linear momentum of the block i.e., J = mv0, which gives
v0 = J/m = 1/0.4 = 2.5 m/s.
The velocity of the particle is given as
$v(t) = {v_0}{e^{ - t/\tau }}$.
Integrate to get the displacement
$x(t) = \int_0^t {{v_0}{e^{ - t/\tau }}dt = {v_0}\tau (} 1 - {e^{ - t/\tau }})$.
Substitute t = $\tau$ = 4 s and v0 = 2.5 m/s to get x($\tau$) = (2.5) (4) (1 $-$ e$-$1) = 6.3 m.
Explanation:
The pushing force ${F_1} = mg\sin \theta + f$
$\therefore$ ${F_1} = mg\sin \theta + \mu mg\cos \theta = mg(\sin \theta + \mu \cos \theta )$
The force required to just prevent it from sliding down
${F_2} = mg\sin \theta - \mu N = mg(\sin \theta - \mu \cos \theta )$
Given, ${F_1} = 3{F_2}$
$\therefore$ $\sin \theta + \mu \cos \theta = 3(\sin \theta - \mu \cos \theta )$
$\therefore$ $1 + \mu = 3(1 - \mu )$ [$\because$ $\sin \theta = \mu \cos \theta $]
$\therefore$ $4\mu = 2$
$\therefore$ $\mu = 0.5$
$\therefore$ $N = 10\mu = 5$
A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is $\mu=\frac{2}{5}$. The disc has an acceleration of $25 \mathrm{~m} / \mathrm{s}^2$. Find the acceleration of the block with respect to disc.
Explanation:
Horizontal disc has a normal reaction $N_1=m g$. The normal reaction $\mathrm{N}_2=m g \sin \theta$. acceleration of the man with respect to the disc is $a^{\prime}$.
$ \begin{aligned} a^{\prime} & =\frac{\mathrm{R}}{m}=a \cos \theta-\mu_k \mathrm{~g}-\mu_k a \sin \theta \\ a^{\prime} & =25\left(\frac{4}{5}\right)-\left(\frac{2}{5}\right)(10)-\left(\frac{2}{5}\right)(25)\left(\frac{3}{5}\right) \\ a^{\prime} & =20-4-6 \\ a^{\prime} & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $
OR
Method (II)
Normal reaction in vertical direction $\mathrm{N}_1=m g$
$\Rightarrow$ Normal reaction from side to the groove
$ \mathrm{N}_2=\mathrm{ma} \sin 37^{\circ} $
Therefore, acceleration of block with respect to disc.
$ \begin{aligned} a r & =\frac{m a \cos 37^{\circ}-\mu \mathrm{N}_1-\mu \mathrm{N}_2}{m} \\ & =\frac{m a \cos 37^{\circ}-\mu m g-\mu m a \sin 37^{\circ}}{m} \\ a r & =\frac{1 \times 25 \times \frac{4}{5}-\frac{2}{5} \times 1 \times 10-\frac{2}{5} \times \frac{3}{5} \times 1 \times 25}{1} \\ a r & =20-4-6 \\ a r & =10 \mathrm{~m} / \mathrm{s}^2 . \end{aligned} $
