$A$ force $\overrightarrow F = (5\hat i + 4\hat j) \ N$ acts on a body and produces a displacement $\overrightarrow S = (6\hat i - 5\hat j + 3\hat k) \ m$. The work done will be $J$.

Consider a particle on which constant forces $F_{1}=\hat{i}+2 \hat{j}+3 \hat{k} \text{ N}$ and $F_{2}=4 \hat{i}-5 \hat{j}-2 \hat{k} \text{ N}$ act together, resulting in a displacement from position $r_{1}=20 \hat{i}+15 \hat{j} \text{ cm}$ to $r_{2}=7 \hat{k} \text{ cm}$. The total work done on the particle is:

If a force $\vec{F}=(3 \hat{i}-2 \hat{j}) \text{ N}$ acting on a body displaces it from point $(1 \text{ m}, 2 \text{ m})$ to point $(2 \text{ m}, 0 \text{ m})$,then the work done by the force is (in $\text{ J}$)

$A$ body of mass $2 \ kg$ is moving along the $x$-direction such that its displacement as a function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma \ m$, where $\alpha = 1 \ m/s^2$, $\beta = 1 \ m/s$, and $\gamma = 1 \ m$. The work done on the body during the time interval $t = 2 \ s$ to $t = 3 \ s$ is . . . . . . $J$.

What is required for work to be done when a force is applied to an object?

Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by