$A$ man of weight $W$ is standing on a lift which is moving downward with an acceleration $a$. The apparent weight of the man is

  • A
    $W(1 + \frac{a}{g})$
  • B
    $W$
  • C
    $W(1 - \frac{a}{g})$
  • D
    $W(1 - \frac{a^2}{g^2})$

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$A$ block is lying at rest inside a bus. What is the maximum acceleration of the bus such that the block remains stationary? (The static friction coefficient $\mu = 0.2$,acceleration due to gravity $g = 10 \ m/s^2$)

$A$ frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. $A$ coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-inertial frame of reference. The relationship between the force $\vec{F}_{\text{rot}}$ experienced by a particle of mass $m$ moving on the rotating disc and the force $\vec{F}_{\text{in}}$ experienced by the particle in an inertial frame of reference is $\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega}$,where $\vec{v}_{\text{rot}}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc,the $x$-axis along the slot,the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega} = \omega \hat{k})$. $A$ small block of mass $m$ is gently placed in the slot at $\vec{r} = (R/2) \hat{i}$ at $t = 0$ and is constrained to move only along the slot.
$(1)$ The distance $r$ of the block at time $t$ is:
$(A)$ $\frac{R}{4}(e^{\omega t} + e^{-\omega t})$
$(B)$ $\frac{R}{2} \cos \omega t$
$(C)$ $\frac{R}{4}(e^{2\omega t} + e^{-2\omega t})$
$(D)$ $\frac{R}{2} \cos 2\omega t$
$(2)$ The net reaction of the disc on the block is:
$(A)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} - e^{-\omega t}) \hat{j} + mg \hat{k}$
$(B)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} + e^{-\omega t}) \hat{j} + mg \hat{k}$
$(C)$ $-m \omega^2 R \cos \omega t \hat{j} - mg \hat{k}$
$(D)$ $m \omega^2 R \sin \omega t \hat{j} - mg \hat{k}$

$A$ child stands on a weighing machine inside a lift. When the lift is going down with acceleration $\frac{g}{3}$,the machine shows a reading $20 \ N$. When the lift goes upwards with acceleration $\frac{g}{3}$,the reading would be ($g=$ gravitational acceleration). (in $N$)

In the figure shown,$P$ is a plate on which a wedge $B$ is placed,and on $B$ a block $A$ of mass $m$ is placed. The plate is suddenly removed and the system of $B$ and $A$ is allowed to fall under gravity. Neglecting any force due to air on $A$ and $B$,the normal force on $A$ due to $B$ is

$A$ man of weight $mg$ is in a lift moving up with an acceleration of $4g$. The apparent weight of the man in the lift is:

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