$A$ body takes just twice the time to slide down a plane inclined at $30^\circ$ to the horizontal as it would if the plane were frictionless. The coefficient of friction between the body and the plane is:

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$ with the horizontal,it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object is shot with the same velocity,it can travel $x_{2}$ distance. Then $x_{1}: x_{2}$ will be

$A$ 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 is projected upwards on an inclined plane of inclination $37^{\circ}$ along the line of greatest slope with a coefficient of friction $\mu = 0.5$ and an initial velocity of $5 \ m/s$. The block first stops at a distance of .......... $m$ from the starting point.

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)