The total work done on a particle is equal to the change in its kinetic energy. This is applicable:

An open knife edge of mass $50 \ g$ is dropped from a height $2 \ m$ on a wooden floor. If the blade penetrates up to a depth of $10 \ cm$ into the wood,the average resistance offered by the wood to the knife edge is $....N$

$A$ body of mass $2 \, kg$ is thrown vertically upwards with a velocity of $4 \, m/s$. At what height will its kinetic energy be half of its initial value (in $, m$)? (Take $g = 10 \, m/s^2$)

$A$ uniform chain of mass $M$ and length $L$ is released from rest from a smooth horizontal surface as shown in the figure. The velocity of the chain at the instant it completely comes off the table will be:

$A$ truck accelerates from speed $v$ to $2v$. The work done during this process 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 $g = 10 \ m/s^2$):