$A$ machine with efficiency $2/3$ uses $12 \text{ J}$ of energy to lift a $2 \text{ kg}$ block to a certain height. If the block is allowed to fall freely from that height,what is its velocity when it reaches the ground?

The potential energy in joules of a particle of mass $1\, kg$ moving in a plane is given by $U = 3x + 4y$,where the position coordinates $x$ and $y$ are measured in metres. If the particle is initially at rest at $(6, 4)$,then:

An object of mass $m$ is sliding down a hill of arbitrary shape and after traveling a certain horizontal path stops because of friction. The friction coefficient is different for different segments of the entire path but it is independent of the velocity and direction of motion. The work done that a force must perform to return the object to its initial position along the same path would be:

$A$ bullet of $10\, \text{g}$, moving with velocity $v$, collides head-on with the stationary bob of a pendulum and recoils with velocity $100\, \text{m/s}$. The length of the pendulum is $0.5\, \text{m}$ and the mass of the bob is $1\, \text{kg}$. The minimum value of $v$ in $\text{m/s}$ so that the pendulum completes a vertical circle is: (Assume the string to be inextensible and $g=10\, \text{m/s}^2$)

Two identical blocks $A$ and $B$, each of mass $m$, resting on a smooth floor, are connected by a light spring of natural length $L$ and spring constant $k$. The spring is at its natural length. $A$ third identical block $C$ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. The maximum compression in the spring is proportional to

In two separate collisions,the coefficients of restitution $e_1$ and $e_2$ are in the ratio $3:1$. In the first collision,the relative velocity of approach is twice the relative velocity of separation. Find the ratio between the relative velocity of approach and the relative velocity of separation in the second collision.