Friction Multiple Choice questions

$v=\frac{Ag}{W\omega}sin{\omega}t$ $\frac{dx}{dt}=\frac{Ag}{W\omega}sin{\omega}t $
or,
$dx=\frac{Ag}{W\omega}sin{\omega}tdt$
Integrating and taking x = 0 at t = 0
$x=\frac{Ag}{W\omega^2}(1-cos{\omega}t)$
hence (b) is the correct option

Time period of oscillation is still is
$T_0=2\pi\sqrt{\frac{l}{g}}$
$T=2\pi\sqrt{\frac{l}{g\pm a}}$
When elevator moves up or down with constant velocity acceleration of the pendulum a=0 and hence $T=T_0$
When the lift is moving upwards with constant acceleration then the weight W=mg is also moving upwards so force equation is
$F_t-W=ma$
$F_t =ma+W=m(a+g)$
Here in this case apparent weight of the pendulum becomes greater than its true weight due to which T becomes less than $T_0$
Where $F_t$ is the tension in the string and W=mg is the weight of the pendulum
When the lift is moving downwards with constant acceleration then the weight W is also moving upwards so force equation is
$W-F_t=ma$
$F_t =W-ma=m(g-a)$
Where $F_t$ is the tension in the string
Here in this case apparent weight of the pendulum becomes less then its true weight due to which T becomes greater than $T_0$
Hence (a) -> (P);(b) -> (P);(c) -> (R);(d) -> (Q)

Answers are self explanatory
(a) -> (Q), (R)
(b) -> (Q), (R)
(c) -> (P), (S)
(d) -> (P), (S)

for system (A)
$m_2g - T = m_2a$
$T = m_1a$
So $m_2g - m_1a = m_2a$
$a = \frac {m_2g}{m_1 + m_2}$
for system (B)
$m_2g - 2T = m_2a$
$T = m_1(2a)$
$m_2g - 4m_1a = m_2a$
$a = \frac {m_2g}{4m_1 +m_2}$
for system (C)
$m_2g - T = m_2a$
$T/2 = 2m_1a$
$T = 4m_1a$
So, $a = \frac {m_2g}{4m_1 +m_2}$
Hence
(a) => (Q) ; (b) => (R) ; (c) => (P) ; (d) => (P)

Answer is (a)
Maximum Value of $f_1 = \mu m_1g=14$ N
Maximum Value of $f_2 = \mu (m_1+m_2)g=10$ N
Maximum Value of $f_3 = \mu (m_1+m_2+m_3)g=9$ N

Answer is (a)
For relative motion between A & B
$F > \mu m_1g$
$F > 0.7 \times 2 \times 10 >14N$
For relative motion between B & C
Force on B must
$F > \mu (m_1+m_2)g$
$F> 10N$
For relative motion between C & floor
Force on C must
$F > \mu (m_1+m_2+m_3)g$
F > 9 N
So at F=9N, all objects will start moving with respect to floor

Answer is (b)
Acceleration will be same for A and B as the F=12N are less then Maximum Frictional force between A and B. So block A and B will move together. Since Force is greater then the frictional force between B and C, there will be relative motion between them
Let write Force equation for each block separately.
For Block A
$12 - f_1 = 2a$
For Block B
$f_1 - f_2 = 3a$

For Block C
$f2 - 9 = 4a_1$
if $a_1 \ne a$
then $f_2= 10$ N
$10 - 9 = 4a_1$
$a_1 = 0.25 \ m/s^2$
$12 - 10 = 5a$
$2 = 5a$
$a = 0.4 \ m/s^2$
So, $f1 = 11.2$ N

Ans. (c)
Force equation for Block C
$f_2 - 9 = 4a_1$
for relative motion between B & C
$10 - 9 = 4a_1$
$a = 0.25 \ m/s^2$

Force equation for Block B
$ f_1 - 10 = 3 \times 0.25$
$f_1 = 10.75 \ N$

Force equation for block A
$F - 10.75 = 0.5$
$F = 11.25 \ N$

Other method
At 9 N all block will start moves with same acceleration a
So $F - f_1 = 2a$
$f_1 - f_2 = 3a$
$f_2 - 9 = 4a$
$ \frac {F - 9}{9} = a$
$f_2 = 4a + 9$
$f_2 = 4 [\frac {(F - 9)}{9}] + 9$

Now $f_2 \leq 10$
$(4F/9) - 4 + 9 \leq 10$
$4F \leq 9 \times 5$
$4F \leq 45$
F = 11.25

Force equation of Block A
$F - f_1 = 2a$
Force equation of Block B
$f_1 - f_2 = 3a$
Force equation of Block C
$f_2 - 9 = 4a_1$
Now for relative motion between B & C
$10 - 9 = 4a_1$
$a_1 = 0.25 \ m/s^2$
Now for relative motion between A & B, $f_1=14N$, So
$F - 14 = 2a$ for Block A
$14 - 10 = 3a$ for Block B
$F - 14 = 2 \times 4/3$
F = 14 + 2.6
F = 16.6 N

Correct option are (b) and (c)
It is explained from above two questions

Force equation for pulley
2T=F or T=F/2 ----(1)
Force equation for Mass
T = ma ---(2)
Solving equation (1) and (2),we get
a = F/2m

Force equation for pulley
2T=F or T=F/2 ----(1)
Force equation for Mass
T -F/4 = ma ---(2)
Substituting the value of T from equation (1) in equation (2)
(F/2) - (F/4) = ma
F/4 = ma
a = F/ 4m

Force of the charge will be given by F=QE
Now It is given
E=-Ej
So , F_x = 0, F_y = -QE j
Hence (a) is the correct option

From above question we know that
F=-QE j
Or ma=-QE j
a=-(QE/m) j
Now velocity equation
v=u+at
So
v= v0i - (qEt/M) j
Hence (c) is the correct option

$\mathbf{r}=\mathbf{v_0}t+\frac{1}{2}\mathbf{a} t^2$
Substituting the values from above
$\mathbf{r}=v_0t\mathbf{i}-\frac{qEt^2}{2M}\mathbf{j}$

Equation of force for Block
Ma=F-f --(1)
Equation of force for Body
ma=f ---(2)
Eliminating a from equation (1) and (2),we get
f=mF/(m+M)
Now $f \leq kmg$
Or
$F \leq k(M+m)g$

Equation of force for Block
Mb=F-f --(1)
Equation of force for Body
ma=f ---(2)
When relative motion is there
f=kmg
or
Mb=F-kmg
ma=kmg
or
$a=kg$
$b=\frac{F-kmg}{M}$

From above question we know,
a=kg
$b=\frac{F-kmg}{M}$
It is obvious b > a
The acceleration of the body relative to block will be directed opposite to the motion and will be equal in magnitude
=(b-a)
$=\frac{F-kmg}{M}-kg$
Now
$L=\frac{1}{2}at^2$
Or $t=\sqrt{\frac{2LM}{F-kg(M+m)}}$

Car has uniform horizontal velocity
$\frac{dx}{dt}=v$
So we can say x=vt
Now
$v_y=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{A\pi v}{\lambda}cos{\left(\frac{2\pi x}{\lambda}\right)}$

Now vertical acceleration
$a_y=\frac{dv_y}{dt}=-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi x}{\lambda}\right)}$
$a_y=-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi vt}{\lambda}\right)}$
So mass m is in accelerated frame of reference, with respect to this frame, the apparent weight of mass is
$W=m(g+a_y)$
$W=m\left[g-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi vt}{\lambda}\right)}\right]$