When a body moves in a circular path,no work is done by the centripetal force since

$A$ body of mass $m$ moves in a circle of radius $r$ with a constant speed $V$. $A$ force of $mV^2/r$ acts towards the center of the circle. Find the work done by this force when the body covers half the circumference of the circle.

$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec{F} = (3\hat{i} + 4\hat{j})$ through two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

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 $\vec F = (5\hat i + 3\hat j) \text{ N}$ is applied to a particle,which displaces it from its origin to the point $\vec r = (2\hat i - 1\hat j) \text{ m}$. The work done on the particle is ........... $J$.