$A$ mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is

The velocity-time graph for a body of mass $10 \ kg$ is shown in the figure. The work done on the body in the first two seconds of motion is:

$A$ particle of mass $m \ kg$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to

$A$ car comes to rest after traveling a distance $s$ under a constant retarding force $F$. If the mass of the car is increased by $50\%$,at what distance will the car come to rest,assuming the same retarding force $F$ and initial velocity $v$?

$A$ particle is released from height $S$ above the surface of the earth. At a certain height,its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:

$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$ as shown in the figure. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ............. $m/s$.