$A$ car is going at a speed of $6\, m/s$ when it encounters a $15\, m$ slope of angle $30^o$. The friction coefficient between the road and tyre is $0.5$. The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ........ $m/s$. $(g = 10\, m/s^2)$

$A$ block of mass $5 \text{ kg}$ starts up a $45^{\circ}$ inclined plane with an initial kinetic energy of $100 \text{ J}$. If the coefficient of friction between the block and the plane is $0.5$,then the distance covered by the block before it stops is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

$A$ block of weight $1 \,N$ rests on an inclined plane of inclination $\theta$ with the horizontal. The coefficient of friction is $\mu$. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

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

$A$ block of mass $M$ is placed on a rough inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$. $A$ force $F$ is applied parallel to the inclined plane as shown in the figure,such that the block just starts moving upward. The value of $F$ is: