$A$ block of mass $m$ slides down a plane inclined at an angle $\theta$. Which of the following will $NOT$ increase the energy lost by the block due to friction?

Consider a uniform cubical box of side $a$ on a rough floor that is to be moved by applying a minimum possible force $F$ at a point $b$ above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$,the maximum possible value of $100 \times \frac{b}{a}$ for a box not to topple before moving is

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \, m/s$ stops after $0.5 \, s$,then the coefficient of friction will be nearly:

If for an inclined plane the coefficient of static friction is $\mu_s = \frac{3}{4}$,then the angle of repose for the inclined plane will be ........ $^o$.

$A$ rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
$(a)$ At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
$(b)$ What is the force acting on the box down the plane,if the angle of inclination of the plane is increased to $\alpha > \theta$?
$(c)$ What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
$(d)$ What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$?