Q426
Allen
Newtons Laws of Motion
Numerical
Calculate the relative acceleration of A w.r.t. B if B is moving with acceleration $a_{0}$ towards right.
Correct Answer: $\frac{a_{0}}{2}$
Explanation:
$\mathrm{Ans.} \frac {a _ {0}}{2}$
$- 3 T a _ {B} + 2 T a _ {1} = 0$
$a _ {1} = \frac {3 a _ {B}}{2} = \frac {3 a _ {0}}{2}$
$a _ {A / B} = \frac {3 a _ {0}}{2} - a _ {0} = \frac {a _ {0}}{2}$
$- 3 T a _ {B} + 2 T a _ {1} = 0$
$a _ {1} = \frac {3 a _ {B}}{2} = \frac {3 a _ {0}}{2}$
$a _ {A / B} = \frac {3 a _ {0}}{2} - a _ {0} = \frac {a _ {0}}{2}$
Q427
Allen
Newtons Laws of Motion
Numerical
In the figure shown all strings and pulleys are massless. If all surfaces are smooth and mass of the blocks are M = 15 kg and m = 10 kg then find the acceleration of mass m (in $m/s^{2}$ ) (Given angle of incline $\theta = 30^{\circ}$ )
Correct Answer: 4
Explanation:
Ans. 4
1 5 g - 2 T = 1 5 a
$2 T - 1 0 g \sin 3 0 ^ {\circ} = 1 0 a$
$a = 4 m / s ^ {2}$
1 5 g - 2 T = 1 5 a
$2 T - 1 0 g \sin 3 0 ^ {\circ} = 1 0 a$
$a = 4 m / s ^ {2}$
Q428
Allen
Newtons Laws of Motion
Numerical
A box of mass m is placed on a smooth horizontal platform as shown in the figure. The platform is made to move in direction $30^{\circ}$ above the horizontal with acceleration a so that the contact force between the box and the platform becomes 3mg/2. Find the magnitude of the acceleration ( $m/s^{2}$ ).
Correct Answer: 10
Explanation:
Ans. 10
$N - m g = \frac {m a}{2}$
$\frac {3 m g}{2} - m g = \frac {m a}{2}$
a = g
$N - m g = \frac {m a}{2}$
$\frac {3 m g}{2} - m g = \frac {m a}{2}$
a = g
Q429
Allen
Newtons Laws of Motion
Numerical
In the arrangement shown $m_{A} = 1kg$ and $m_{B} = 2kg$ while all the pulleys and strings are massless and frictionless. At t = 0, a force F = 10t starts acting over central pulley in vertical direction. Where B losses contact with floor the velocity of A is (5n) m/s. Find out the value of 'n'.
Correct Answer: 2
Explanation:
Ans. 2
Block A leave contact at t = 2 sec and Block B leave contact at t = 4 sec.
$a _ {A} = 5 t - 1 0$
$v = \int_ {2} ^ {4} a _ {A}. d t$
$\begin{array}{l} \int_ {2} ^ {4} (5 t - 1 0) d t \\ = 1 0 m / \sec \Rightarrow 1 0 = 5 n \\ n = 2 \end{array}$
Block A leave contact at t = 2 sec and Block B leave contact at t = 4 sec.
$a _ {A} = 5 t - 1 0$
$v = \int_ {2} ^ {4} a _ {A}. d t$
$\begin{array}{l} \int_ {2} ^ {4} (5 t - 1 0) d t \\ = 1 0 m / \sec \Rightarrow 1 0 = 5 n \\ n = 2 \end{array}$
Q430
Allen
Newtons Laws of Motion
Numerical
Velocities of blocks A, B and pulley $p_{2}$ are shown in figure. Find velocity of pulley $p_{1}$ and block C.
Correct Answer: $V_{P_{1}} = 5 \, m/s$ downward, $V_{C} = 25 \, m/s$ upward
Explanation:
Ans. $V_{P_{1}} = 5 \, m/s$ downward, $V_{C} = 25 \, m/s$ upward
$V _ {1} = \frac {1 0 - 2 0}{2}$
[Constrained relation of $P_{1}$ ]
$V _ {1} = - 5 m / s$
$1 0 = \frac {- 5 + V _ {2}}{2}$
$V _ {2} = 2 5 m / s \uparrow$
$V_{C} = V_{2} = 25m / s$ upwards
$V _ {P _ {1}} = V _ {1} = - 5 m / s$
$\Rightarrow V_{P} = 5m / s$ downward
[Because we have assumed upward direction as +ve for $V_{1}$ ]
$V _ {1} = \frac {1 0 - 2 0}{2}$
[Constrained relation of $P_{1}$ ]
$V _ {1} = - 5 m / s$
$1 0 = \frac {- 5 + V _ {2}}{2}$
$V _ {2} = 2 5 m / s \uparrow$
$V_{C} = V_{2} = 25m / s$ upwards
$V _ {P _ {1}} = V _ {1} = - 5 m / s$
$\Rightarrow V_{P} = 5m / s$ downward
[Because we have assumed upward direction as +ve for $V_{1}$ ]
Q431
Allen
Newtons Laws of Motion
Numerical
Find the acceleration of rod A and wedge B in the arrangement shown in figure if the mass of rod is equal to that of the wedge and the friction between all contact surfaces is negligible and rod A is free to move downwards only. Take angle of wedge as $37^{\circ}$ .
Correct Answer: $a_{A} = \frac{9g}{25}$ , $a_{B} = \frac{12g}{25}$
Explanation:
Ans. $a_{A} = \frac{9g}{25}, a_{B} = \frac{12g}{25}$
$\tan 37^{\circ} = \frac{a_{A}}{a_{B}}$
(wedge constrained relation)
$N\sin 37^{\circ} = ma_{B}$
For Rod $A \rightarrow mg - N \cos 37^{\circ} = ma_{A}$
From equation (1) and (2) is
$a _ {A} = \frac {9 g}{2 5}, a _ {B} = \frac {1 2 g}{2 5}$
$\tan 37^{\circ} = \frac{a_{A}}{a_{B}}$
(wedge constrained relation)
$N\sin 37^{\circ} = ma_{B}$
For Rod $A \rightarrow mg - N \cos 37^{\circ} = ma_{A}$
From equation (1) and (2) is
$a _ {A} = \frac {9 g}{2 5}, a _ {B} = \frac {1 2 g}{2 5}$
Q432
Allen
Newtons Laws of Motion
Numerical
A rod of mass 2m moves vertically downward on the surface of wedge of mass as shown in figure. Find the relation between velocity of rod and that of the wedge at any instant.
Correct Answer: $u = v \tan \theta$ (u is velocity of rod, v is the velocity of wedge)
Explanation:
Ans. $u = v \tan \theta$ (u is velocity of rod, v is the velocity of wedge)
Using wedge constraint.
Component of velocity of rod along perpendicular to inclined surface is equal to velocity of wedge along that direction.
$u\cos \theta = v\sin \theta$
${\frac {u}{v}} = \tan \theta$
$u = v \tan \theta$
Using wedge constraint.
Component of velocity of rod along perpendicular to inclined surface is equal to velocity of wedge along that direction.
$u\cos \theta = v\sin \theta$
${\frac {u}{v}} = \tan \theta$
$u = v \tan \theta$
Q433
Allen
Newtons Laws of Motion
Numerical
The velocity of end 'A' of rigid rod placed between two smooth vertical walls is 'u' along vertical direction. Find out the velocity of end 'B' of that rod, rod always remains in contact with the vertical wall.
Correct Answer: $u \tan \theta$
Explanation:
Ans. $u\tan \theta$
Since rod is rigid, its length can't increase.
ā“ Velocity of approach of A and B point of rod is zero.
$\Rightarrow u\sin \theta -v\cos \theta = 0$
$\Rightarrow \quad v = u\tan \theta$
Perpendicular to contact of two blocks
Since rod is rigid, its length can't increase.
ā“ Velocity of approach of A and B point of rod is zero.
$\Rightarrow u\sin \theta -v\cos \theta = 0$
$\Rightarrow \quad v = u\tan \theta$
Perpendicular to contact of two blocks
Q434
Allen
Newtons Laws of Motion
Numerical
Find the acceleration of A in term of b.
Correct Answer: $b\hat{i} + b\hat{j}$
Explanation:
Ans. $b\hat{i} + b\hat{j}$
$\ell_ {1} + \ell_ {2} = C$
$\ell_ {1} ^ {\prime \prime} + \ell_ {2} ^ {\prime \prime} = 0$
b - a = 0, a = b
Acceleration of $A$ , $b\hat{i} + b\hat{j}$ .
$\ell_ {1} + \ell_ {2} = C$
$\ell_ {1} ^ {\prime \prime} + \ell_ {2} ^ {\prime \prime} = 0$
b - a = 0, a = b
Acceleration of $A$ , $b\hat{i} + b\hat{j}$ .
Q435
Allen
Newtons Laws of Motion
Numerical
A 2 kg block A is attached to one end of a light string that passes over an ideal pulley and a 1 kg sleeve B slides down the other part of the string with an acceleration of $5 \, m/s^{2}$ with respect to the string. Find the acceleration of the block, acceleration of sleeve and tension in the string. [q = 10 m/sĀ²] $[ g = 1 0 m / s ^ {2} ]$
Correct Answer: $5\ m/s^{2}$ downwards, $0\ m/s^{2}$ , 10 N
Explanation:
Ans. $5 \, m/s^{2}$ downwards, $0 \, m/s^{2}$ , $10 \, N$
2 0 - T = 2 a
for sleeve,
$1 0 - f = 1 a _ {1}$
$a _ {1} + a = 5$
$a _ {1} = 5 - a$
$\boxed {a = 5 m / s ^ {2} \downarrow}$
2 0 - f = 2 a
1 0 - f = 5 - a
1 0 = 3 a - 5
$\mathrm{a} = 1 5 / 3$
$\mathrm{accel},$
$a _ {1} = 5 - 5 = 0$
$2 0 - T = 2 \times 5$
T = 2 0 - 1 0
2 0 - T = 2 a
for sleeve,
$1 0 - f = 1 a _ {1}$
$a _ {1} + a = 5$
$a _ {1} = 5 - a$
$\boxed {a = 5 m / s ^ {2} \downarrow}$
2 0 - f = 2 a
1 0 - f = 5 - a
1 0 = 3 a - 5
$\mathrm{a} = 1 5 / 3$
$\mathrm{accel},$
$a _ {1} = 5 - 5 = 0$
$2 0 - T = 2 \times 5$
T = 2 0 - 1 0
Q436
Allen
Newtons Laws of Motion
Numerical
For the arrangement shown in figure when the system is released, find the acceleration of wedge. Pulley and string are ideal and friction is absent.
Correct Answer: $a = \frac{mg\sin\theta}{M + 2m(1 - \cos\theta)}$
Explanation:
Ans. $a = \frac{mg \sin \theta}{M + 2m(1 - \cos \theta)}$
Considering block and wedge as a system and using. Newton's law for the system along $x$ -direction
$T = M a + m (a - b \cos \theta)\tag{... (i}$
F.B.D of m
Along the inclined plane
$m g \sin \theta - T = m (b - a \cos \theta)\tag{... (ii}$
Using string constraint equation.
$\ell_ {1} + \ell_ {2} = \mathrm{constant}$
$\frac {d ^ {2} \ell_ {1}}{d t ^ {2}} + \frac {d ^ {2} \ell_ {2}}{d t ^ {2}} = 0$
$b - a = 0\tag{... (iii}$
Solving above equations (i), (ii) and (iii), we get
$a = \frac {m g \sin \theta}{M + 2 m (1 - \cos \theta)}$
Considering block and wedge as a system and using. Newton's law for the system along $x$ -direction
$T = M a + m (a - b \cos \theta)\tag{... (i}$
F.B.D of m
Along the inclined plane
$m g \sin \theta - T = m (b - a \cos \theta)\tag{... (ii}$
Using string constraint equation.
$\ell_ {1} + \ell_ {2} = \mathrm{constant}$
$\frac {d ^ {2} \ell_ {1}}{d t ^ {2}} + \frac {d ^ {2} \ell_ {2}}{d t ^ {2}} = 0$
$b - a = 0\tag{... (iii}$
Solving above equations (i), (ii) and (iii), we get
$a = \frac {m g \sin \theta}{M + 2 m (1 - \cos \theta)}$
Q437
Allen
Newtons Laws of Motion
Numerical
In the arrangement shown in figure the mass of ball is $\eta = 1.8$ times as great as that of rod 2. The length of the latter is l = 100 cm. The masses of the pulleys and the threads, as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. How soon will the ball be opposite the upper end of the rod?
Correct Answer: $t = \sqrt{2\ell(4 + \eta) / 3g(2 - \eta)} = 1.4s$
Explanation:
Ans. $t=\sqrt{2\ell(4+\eta)/3g(2-\eta)}=1.4s$
$T - mg = mw_{1}$ ... (ii)
and $Mg = T' = Mw_{2}$ ... (iii)
but $T = 2T$ ... (iv)
(Because pulley is massless)
From Eqs. (1), (2), (3) and (4)
$w_{1} = \frac{(2M - m)g}{m + 4M} = \frac{(2 - \eta)g}{\eta + 4}$ (in upward direction)
and $w_{2} = \frac{2(2 - \eta)g}{(\eta + 4)}$ (downwards)
From kinematical equation in projection form, we get
$l = \frac {1}{2} \left(w _ {1} + w _ {2}\right) t ^ {2}$
As $w_{1}$ and $w_{2}$ are in the opposite direction. Putting the values of $w_{1}$ and $w_{2}$ , the sought time becomes.
$t = \sqrt {2 l (\eta +) / 3 (2 - \eta) g} = 1. 4 s$
$T - mg = mw_{1}$ ... (ii)
and $Mg = T' = Mw_{2}$ ... (iii)
but $T = 2T$ ... (iv)
(Because pulley is massless)
From Eqs. (1), (2), (3) and (4)
$w_{1} = \frac{(2M - m)g}{m + 4M} = \frac{(2 - \eta)g}{\eta + 4}$ (in upward direction)
and $w_{2} = \frac{2(2 - \eta)g}{(\eta + 4)}$ (downwards)
From kinematical equation in projection form, we get
$l = \frac {1}{2} \left(w _ {1} + w _ {2}\right) t ^ {2}$
As $w_{1}$ and $w_{2}$ are in the opposite direction. Putting the values of $w_{1}$ and $w_{2}$ , the sought time becomes.
$t = \sqrt {2 l (\eta +) / 3 (2 - \eta) g} = 1. 4 s$
Q438
Allen
Friction
Numerical
In the figure, what should be mass m so that block A slides up with a constant velocity?
Correct Answer: 1 kg
Explanation:
Ans. 1 kg
Since v = constant
T = m g
$\mathrm{Forblock},\tag{... (1}$
T = 6 + N
$T = 6 + 0. 5 \times 8$
T = 6 + 4
$\boxed {T = 1 0 N}$
From (1) and (2)
$m g = 1 0\tag{...(2}$
$\boxed {m = 1 k g}$
Since v = constant
T = m g
$\mathrm{Forblock},\tag{... (1}$
T = 6 + N
$T = 6 + 0. 5 \times 8$
T = 6 + 4
$\boxed {T = 1 0 N}$
From (1) and (2)
$m g = 1 0\tag{...(2}$
$\boxed {m = 1 k g}$
Q439
Allen
Friction
Numerical
Find the acceleration of the blocks and magnitude & direction of frictional force between block A and table, if block A is pulled towards left with a force of 50 N.
Correct Answer: $10\hat{i}$
Explanation:
Ans. 10 $^{\hat{i}}$
$f _ {m a x} = \mu \times N = 0. 8 \times 5 0 = 4 0 N$
But required friction = 10N
$f = 1 0 \hat {i}$
$f _ {m a x} = \mu \times N = 0. 8 \times 5 0 = 4 0 N$
But required friction = 10N
$f = 1 0 \hat {i}$
Q440
Allen
Friction
Numerical
A block of mass m lies on wedge of mass M as shown in figure. Find the minimum friction coefficient required between wedge M and ground so that it does not move while block m slips down on it.
Correct Answer: $\mu_{min}=\frac{m\sin\theta\cos\theta}{m\cos^{2}\theta+M}$
Explanation:
Ans. $\mu_{min}=\frac{m\sin\theta\cos\theta}{m\cos^{2}\theta+M}$
$N _ {0} = M g + N \cos \theta = M g + m g \cos^ {2} \theta$
$\text { Friction } \mu N _ {0} = \mu (M g + m g \cos^ {2} \theta) = N \sin \theta$
$\mu_ {\min} = \frac {m g \sin \theta \cos \theta}{(M g + m g \cos^ {2} \theta)}$
$\mu_ {\mathrm{min}} = \frac {m \sin \theta \cos \theta}{m \cos^ {2} \theta + M}$
$N _ {0} = M g + N \cos \theta = M g + m g \cos^ {2} \theta$
$\text { Friction } \mu N _ {0} = \mu (M g + m g \cos^ {2} \theta) = N \sin \theta$
$\mu_ {\min} = \frac {m g \sin \theta \cos \theta}{(M g + m g \cos^ {2} \theta)}$
$\mu_ {\mathrm{min}} = \frac {m \sin \theta \cos \theta}{m \cos^ {2} \theta + M}$
Q441
Allen
Friction
Numerical
You pour a quantity of flour of volume $V = 72 \, cm^{3}$ onto a board, where it forms a conical pile. The coefficient of static friction between flour grains is $\mu_{s} = \sqrt{\pi}$ . Find the maximum height (in cm) of the pile.
Correct Answer: 6
Explanation:
Ans. 6
$\tan \theta = \frac {h}{r} = \mu_ {s} \Rightarrow r = \frac {h}{\mu_ {s}}$
$v = \frac {1}{3} \pi r ^ {2} h$
$7 2 = \frac {1}{3} \times \pi \times \frac {h ^ {2}}{\mu_ {s} ^ {2}} \times h = \frac {1}{3} \times \pi \times \frac {h ^ {2}}{\pi} \times h.$
$\Rightarrow h ^ {3} = 2 1 6 \Rightarrow h = 6 c m$
$\tan \theta = \frac {h}{r} = \mu_ {s} \Rightarrow r = \frac {h}{\mu_ {s}}$
$v = \frac {1}{3} \pi r ^ {2} h$
$7 2 = \frac {1}{3} \times \pi \times \frac {h ^ {2}}{\mu_ {s} ^ {2}} \times h = \frac {1}{3} \times \pi \times \frac {h ^ {2}}{\pi} \times h.$
$\Rightarrow h ^ {3} = 2 1 6 \Rightarrow h = 6 c m$
Q442
Allen
Friction
Numerical
Block $M$ slides down on frictionless incline as shown. The minimum friction coefficient so that $m$ does not slide with respect to $M$ is $\frac{n}{4}$ , find the value of $n$ .
Correct Answer: 3
Explanation:
Ans. 3
If $m$ not sliding
$a = g \sin \theta$
$m g - N = m g \sin^ {2} \theta$
$N = m g (1 - \sin^ {2} \theta) = m g \cos^ {2} \theta$
$\mu N = m g \sin \theta \cos \theta$
$\mu = \frac {\sin \theta}{\cos \theta} = \tan \theta = \frac {3}{4}$
If $m$ not sliding
$a = g \sin \theta$
$m g - N = m g \sin^ {2} \theta$
$N = m g (1 - \sin^ {2} \theta) = m g \cos^ {2} \theta$
$\mu N = m g \sin \theta \cos \theta$
$\mu = \frac {\sin \theta}{\cos \theta} = \tan \theta = \frac {3}{4}$
Q443
Allen
Friction
Numerical
A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the wall and the surface of the block is 1/4. Calculate frictional force acting on the block.
Correct Answer: $20N$ vertically downward
Explanation:
Ans. 20 N vertically downward
$f _ {\max} = \mu N = \frac {1}{4} \times 5 0 \sqrt {3} = 1 2. 5 \sqrt {3} = 2 1. 2 5 N$
5 0 = 3 0 + f = f = 2 0 0 < f _ {\max}
Hence $\boxed{f=20N}$ down words
$f _ {\max} = \mu N = \frac {1}{4} \times 5 0 \sqrt {3} = 1 2. 5 \sqrt {3} = 2 1. 2 5 N$
5 0 = 3 0 + f = f = 2 0 0 < f _ {\max}
Hence $\boxed{f=20N}$ down words
Q444
Allen
Friction
Numerical
A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$ . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If define $N = 10\mu$ , then N is
Correct Answer: 5
Explanation:
Ans. 5
$\mu = \frac {\tan \theta}{2}; \mu \frac {1}{2}$
$N = 1 0 \times \frac {1}{2}$
$\mu = \frac {\tan \theta}{2}; \mu \frac {1}{2}$
$N = 1 0 \times \frac {1}{2}$
Q445
Allen
Friction
Numerical
Find minimum value of m so that 10 kg block moves up the inclined plane.
Correct Answer: $20\mathrm{kg}$
Explanation:
Ans. 20 kg
Q446
Allen
Friction
Numerical
A circular disc with a groove along its diameter is placed horizontally. A block of m ass 1 kg is placed as shown. The co-efficient of friction between the block and all surface of groove in contact is $\mu = 2/5$ . The disc has an acceleration of $25 \, m/s^{2}$ . Find the acceleration of the block with respect to disc.
Correct Answer: $10m / s^2$
Explanation:
Ans. $10m / s^2$
$ma \cos 37 - \mu ma \sin 37 - \mu mg = ma_{0}$
$2 5 \times \frac {4}{5} - \frac {2}{5} \times 2 5 \times \frac {3}{5} - \frac {2}{5} \times 1 0 = a _ {0}$
2 0 - 6 - 4 = 9 0
$a _ {0} = 1 0 m / s ^ {2}$
$ma \cos 37 - \mu ma \sin 37 - \mu mg = ma_{0}$
$2 5 \times \frac {4}{5} - \frac {2}{5} \times 2 5 \times \frac {3}{5} - \frac {2}{5} \times 1 0 = a _ {0}$
2 0 - 6 - 4 = 9 0
$a _ {0} = 1 0 m / s ^ {2}$
Q447
Allen
Friction
Numerical
In the arrangement shown coefficient of friction between the two blocks A and B is 0.6 and ground is frictionless. Force constant of spring is 900 N/m. Find the maximum elongation of spring (in cm) such that the blocks do not slip over each other when released.
Correct Answer: 1
Explanation:
Ans. 1
When released $a_{\mathrm{C}} = \frac{kx}{3kg}$ for no slipping $f \leq \mu N$
$f = \frac {2 k x}{3} \leq 0. 6 \times 1 0 \Rightarrow k x \leq 9 \Rightarrow x \leq \frac {9}{9 0 0}$
$\Rightarrow x \leq 0.01m \Rightarrow x \leq 1cm$
When released $a_{\mathrm{C}} = \frac{kx}{3kg}$ for no slipping $f \leq \mu N$
$f = \frac {2 k x}{3} \leq 0. 6 \times 1 0 \Rightarrow k x \leq 9 \Rightarrow x \leq \frac {9}{9 0 0}$
$\Rightarrow x \leq 0.01m \Rightarrow x \leq 1cm$
Q448
Allen
Friction
Numerical
Coefficient of friction between 5 kg and 10 kg block is 0.5. If friction between them is 20 N. What is the value of force being applied on 5 kg. The floor is frictionless.
Correct Answer: $30N$
Explanation:
Ans. $30N$
$f _ {\max} = \mu N = 0. 5 \times 5 0 = 2 5 N$
$f < f_{nel}$ hence acceleration of both blocks are same
F - f = 5 a
f = 1 0 a
2 0 = 1 0 a
a = 2
F - 2 0 = 1 0
F = 3 0 N
$f _ {\max} = \mu N = 0. 5 \times 5 0 = 2 5 N$
$f < f_{nel}$ hence acceleration of both blocks are same
F - f = 5 a
f = 1 0 a
2 0 = 1 0 a
a = 2
F - 2 0 = 1 0
F = 3 0 N
Q449
Allen
Friction
Numerical
In the figure masses $m_{1}$ , $m_{2}$ and M are 20 kg, 5 kg and 50 kg respectively. The co-efficient of friction between M and ground is zero. The co-efficient of friction between $m_{1}$ and M and that between $m_{2}$ and ground is 0.3. The pulleys and the string are massless. The string is perfectly horizontal between $P_{1}$ and $m_{1}$ and also between $P_{2}$ and $m_{2}$ . The string is perfectly vertical between $P_{1}$ and $P_{2}$ . An external horizontal force F is applied to the mass M. Take $g = 10 \, m/s^{2}$ .
(I) Draw a free-body diagram for mass M, clearly showing all the forces.
(II) Let the magnitude of the force of friction between $m_{1}$ and M be $f_{1}$ and that between $m_{2}$ and ground be $f_{2}$ . For a particular F it is found that $f_{1} = 2f_{2}$ . Find $f_{1}$ and $f_{2}$ . Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. 
Correct Answer: ii) 60N
Explanation:
Ans.
(i)
(ii)
$a = 3 / 5 m / s ^ {2}, T = 1 8 N, F = 6 0 N$
$f _ {1} = 2 f _ {2}$
For bigger block
$f _ {1} - T = 2 0 a$
$F - f _ {1} = 5 0 a$
$T - f _ {2} = 5 a$
$F - 2 f _ {2} = 5 0 a$
... (i)
$2 f _ {2} - T = 2 0 a$
... (ii)
$T - f _ {2} = 5 a$
... (iii)
$f _ {2} = 2 5 a \Rightarrow 1 5 = 2 5 a \Rightarrow a = a = \frac {3}{5} m / s$
From equation (i)
$F _ {2} - 5 0 a = 5 0 a$
$F _ {2} = 1 0 0 a = 6 0 N$
$f _ {2} = \mu m _ {2} g = 0. 3 \times 5 0$
= 1 5 N
T - 1 5 = 3
T = 1 8 N
(i)
(ii)
$a = 3 / 5 m / s ^ {2}, T = 1 8 N, F = 6 0 N$
$f _ {1} = 2 f _ {2}$
For bigger block
$f _ {1} - T = 2 0 a$
$F - f _ {1} = 5 0 a$
$T - f _ {2} = 5 a$
$F - 2 f _ {2} = 5 0 a$
... (i)
$2 f _ {2} - T = 2 0 a$
... (ii)
$T - f _ {2} = 5 a$
... (iii)
$f _ {2} = 2 5 a \Rightarrow 1 5 = 2 5 a \Rightarrow a = a = \frac {3}{5} m / s$
From equation (i)
$F _ {2} - 5 0 a = 5 0 a$
$F _ {2} = 1 0 0 a = 6 0 N$
$f _ {2} = \mu m _ {2} g = 0. 3 \times 5 0$
= 1 5 N
T - 1 5 = 3
T = 1 8 N
Q450
Allen
JEE Main PYQ
MCQ
A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic friction is $\mu_{K}$ . Then, the block's acceleration 'a' is given by: (g is acceleration due to gravity)
A.
$-\frac{F}{m}\cos\theta - \mu_{K}\left(g - \frac{F}{m}\sin\theta\right)$
B.
$\frac{F}{m}\cos\theta - \mu_{K}\left(g - \frac{F}{m}\sin\theta\right)$
C.
$\frac{F}{m}\cos\theta - \mu_{K}\left(g + \frac{F}{m}\sin\theta\right)$
D.
$\frac{F}{m}\cos\theta + \mu_{K}\left(g - \frac{F}{m}\sin\theta\right)$