$A$ ball of mass $20 \ kg$ is at rest on a hill of height $100 \ m$. It starts rolling down,reaches the ground,climbs another hill of height $30 \ m$,and finally rolls down to a horizontal platform at a height of $20 \ m$ from the ground. What will be its velocity at this point? (Assume $g = 10 \ m/s^2$)

Consider an elliptically shaped rail $PQ$ in the vertical plane with $OP = 3 \ m$ and $OQ = 4 \ m$. $A$ block of mass $1 \ kg$ is pulled along the rail from $P$ to $Q$ with a force of $18 \ N$,which is always parallel to the line $PQ$ (see the figure). Assuming no frictional losses,the kinetic energy of the block when it reaches $Q$ is $(n \times 10) \ J$. The value of $n$ is (take acceleration due to gravity $g = 10 \ m/s^2$):

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

$A$ body of mass $0.25 \ kg$ travels along a straight line from $x=0$ to $x=2 \ m$ with a speed $v=k x^{3/2}$ where $k=2$ $SI$ units. The work done by the net force during this displacement is: (in $J$)

$A$ balloon filled with helium rises against gravity,increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that the density of air is constant.

$A$ force of $5\,N$,making an angle $\theta$ with the horizontal,acts on an object and displaces it by $0.4\,m$ along the horizontal direction. If the object gains a kinetic energy of $1\,J$,the horizontal component of the force is......$N$. (in $.5$)