$A$ student of mass $M$ is $1.5 \,m$ tall and has her centre of mass $1 \,m$ above the ground when standing straight. She wants to jump up vertically. To do so,she bends her knees so that her centre of mass is lowered by $0.2 \,m$ and then pushes the ground by a constant force $F$. As a result,she jumps up such that the maximum height of her feet is $0.3 \,m$ above the ground. The ratio $F / Mg$ is (in $.5$)

$A$ mass of $2 \ kg$,initially at a height of $1.2 \ m$ above an uncompressed spring with spring constant $2 \times 10^4 \ N/m$,is released from rest to fall on the spring. Taking the acceleration due to gravity as $10 \ m/s^2$ and neglecting air resistance,the compression of the spring in $mm$ is:

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 $g = 10 \ m/s^2$):

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ until the mass comes to rest,as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $A$,the net work done on the block is

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

The mass of a pendulum bob is $50 \ gm$. The bob is released from a horizontal position $A$ as shown in the figure. If the length of the pendulum is $1.5 \ m$,what will be its kinetic energy when it reaches the lowest point $B$ of its path? (Take $g = 10 \ m/s^2$)