$A$ particle is displaced from point $P(3 \ m, 4 \ m, 5 \ m)$ to a point $Q(2 \ m, 3 \ m, 4 \ m)$ under a constant force $\vec{F}=(3 \hat{i}+4 \hat{j}+5 \hat{k}) \ N$. The work done by the force in this process is

$A$ 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

$A$ uniform force of $(3\hat i + \hat j) \text{ N}$ acts on a particle of mass $2 \text{ kg}$. The particle is displaced from position $(2\hat i + \hat k) \text{ m}$ to position $(4\hat i + 3\hat j - \hat k) \text{ m}$. The work done by the force on the particle is ............. $J$.

$A$ 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 $\,J$)

If a force $\vec F = 4\hat i + 5\hat j$ causes a displacement $\vec s = 3\hat i + 6\hat k$,the work done is:

$A$ particle is made to move from the origin in three spells of equal distances,first along the $x$-axis,second parallel to the $y$-axis,and third parallel to the $z$-axis. One of the forces acting on it has a constant magnitude of $50 \ N$ and always acts along the direction of motion. Work done by this force in the three spells of motion are equal,and the total work done in all three spells is $300 \ J$. The final coordinates of the particle will be: