$A$ force of magnitude $50\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a point mass from $(5, 9, 7)$ to $(4, 8, 6)$. The work done by this force during this displacement is:

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 $J$?

$A$ cyclist comes to a skidding stop in $10 \; m$. During this process,the force on the cycle due to the road is $200 \; N$ and is directly opposed to the motion.
$(a)$ How much work does the road do on the cycle?
$(b)$ How much work does the cycle do on the road?

$A$ force acts on a $30 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = \alpha t^2$,where $x$ is in metre,$t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ 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

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