$A$ horizontal force of $5 \,N$ is required to maintain a velocity of $2 \,m/s$ for a block of $10 \,kg$ mass sliding over a rough surface. The work done by this force in one minute is $J$.

$A$ 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 $F = (5\hat{i} + 3\hat{j}) \ N$ acts on an object,causing a displacement of $r = (2\hat{i} - 1\hat{j}) \ m$. Calculate the work done in $J$.

$A$ 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 $\,J$)

The work done against gravity in taking a $10 \, kg$ mass to a $1 \, m$ height in $1 \, s$ will be .... $J$.

The kinetic energy acquired by a body of mass $M$ in travelling a certain distance $d$,starting from rest,under the action of a constant force is: