$A$ body of mass $m$ kg is lifted by a man to a height of one metre in $30\, s$. Another man lifts the same mass to the same height in $60\, s$. The work done by them are in the ratio:

Show that a force that does no work must be a velocity-dependent force.

$A$ body of mass $6 \,kg$ is under a force which causes displacement in it given by $s = \frac{t^2}{4} \,m$, where $t$ is time in seconds. The work done by the force in $2 \,s$ is: (in $\,J$)

$A$ force of $4 \,N$ acts on a $10 \,kg$ body initially at rest. Let $W_1$ be the work done by the force during $0 \leq t \leq 1 \,s$. Likewise, $W_2$ is the work done by the force during $1 \,s \leq t \leq 2 \,s$, where $t$ is time in seconds. The ratio $\frac{W_2}{W_1}$ is

$A$ man carries a bucket of weight $60 \ N$ and walks $7 \ m$ in the horizontal direction,then climbs $5 \ m$ in the vertical direction. What is the work done by the man in $J$?

Write the definition of work and kinetic energy.