$A$ force $F = (2 \hat{i} + 4 \hat{j}) \text{ N}$ is applied on an object of mass $M$. What is the work done by this force in moving the object horizontally along the $X$-axis by $3 \text{ m}$ (in $\text{ J}$)?

$A$ 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 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 $\,J$)

Given that $W = \vec F \cdot \vec s = 0$ and $F \neq 0, s \neq 0$,then

$A$ particle moves from position $\vec{r_1} = 3\,\hat{i} + 2\,\hat{j} - 6\,\hat{k}$ to position $\vec{r_2} = 14\,\hat{i} + 13\,\hat{j} + 9\,\hat{k}$ under the influence of a force $\vec{F} = 4\,\hat{i} + \hat{j} + 3\,\hat{k}$. Find 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 $J$?