A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $\mathrm{v}=\alpha \sqrt{\mathrm{x}}$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $\mathrm{x}=0$ to $\mathrm{x}=\mathrm{d}$, will be:
$\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}$
$\frac{\mathrm{md}}{2 \alpha^2}$
$\frac{m \alpha^2 d}{2}$
$2 m \alpha^2 d$
When work is done on a body by an external force, its
A boy is rolling a $0.5\, kg$ ball on the frictionless floor with the speed of $20\, ms ^{-1}$. The ball gets deflected by an obstacle on the way. After deflection it moves with $5 \%$ of its initial kinetic energy. What is the speed of the ball now ? (in $ms ^{-1}$)
A small bar starts sliding down on inclined plane forming an angle $\theta $ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$ , where $k$ is a constant. Find the distance covered by the bar till it stops
A particle of mass $m$ slides from rest down a plane inclined at $30^o$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$ where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is :-
A spherical ball of mass $20\, kg$ is stationary at the top of a hill of height $100 \,m$. It slides down a smooth surface to the ground, then climbs up another hill of height $30 \,m$ and finally slides down to a horizontal base at a height of $20 \,m$ above the ground. The velocity attained by the ball is ............... $\mathrm{m} / \mathrm{s}$