$A$ body is thrown vertically upwards and reaches a maximum height $h$. What is the ratio of its kinetic energy to its potential energy at a height of $\frac{3h}{4}$?

Explain the conservation of mechanical energy using the example of a falling ball.

$A$ body of mass $m$ slides down a smooth hemispherical surface of radius $r$. What will be its velocity at the bottom of the surface?

$A$ particle,initially at rest on a frictionless horizontal surface,is acted upon by a horizontal force which is constant in size and direction. $A$ graph is plotted between the work done $(W)$ on the particle and the speed of the particle $(v)$. If there are no other horizontal forces acting on the particle,the graph would look like:

$A$ block of mass $\sqrt{2} \, kg$ is released from the top of an inclined smooth surface as shown in the figure. If the spring constant of the spring is $100 \, N/m$ and the block comes to rest after compressing the spring by $1 \, m$,then the total distance travelled by the block along the incline before it comes to rest is ......... $m$.

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is: