$A$ simple pendulum bob of mass $m$ is raised to a maximum height $h$. Find the work done by the gravitational force as the pendulum moves from one extreme end to the other.

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

$A$ man of mass $50 \ kg$ carries a load of mass $20 \ kg$ and climbs $20$ steps,each of height $0.25 \ m$. Calculate the work done in $J$.

Two constant forces ${F_1} = 2\hat i - 3\hat j + 3\hat k$ $(N)$ and ${F_2} = \hat i + \hat j - 2\hat k$ $(N)$ act on a body and displace it from the position ${r_1} = \hat i + 2\hat j - 2\hat k$ $(m)$ to the position ${r_2} = 7\hat i + 10\hat j + 5\hat k$ $(m)$. What is the work done 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

$A$ body is being raised to height $h$ from the surface of the Earth. What is the sign of work done by $(a)$ applied force and $(b)$ gravitational force?