$A$ force of $5 \ N$ is required to maintain a constant velocity of $2 \ m/s$ for a block of mass $10 \ kg$ on a rough surface. The work done by this force in $1 \ minute$ is: (in $J$)

$A$ gardener pushes a lawn roller through a distance of $20 \,m$. If he applies a force of $30 \,kg-wt$ in a direction inclined at $60^{\circ}$ to the ground, the work done by the gardener in pushing the roller is (Given: $g=9.8 \,m/s^2$) (in $\,J$)

$A$ force of magnitude $30\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a particle from point $(2, 4, 1)$ to $(3, 5, 2)$. The work done during this displacement 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:

$A$ body moves a distance of $10 \, m$ along a straight line under the action of a force of $5 \, N$. If the work done is $25 \, J$,the angle which the force makes with the direction of motion of the body is.....$^o$

An adult weighing $600 \, N$ raises the centre of gravity of his body by $0.25 \, m$ while taking each step of $1 \, m$ length in jogging. If he jogs for $6 \, km$,calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $10 \, \%$ of energy intake in the form of food,calculate the energy equivalents of food that would be required to compensate for the energy utilised for jogging.