If a constant force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$ acting on a body of mass $5 \text{ kg}$ displaces it from $(3 \hat{i} - 4 \hat{k}) \text{ m}$ to $(2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$, then the work done by the force on the body is (in $\text{ J}$)

$A$ cyclist comes to a skidding stop in $10 \,m$. During this process,the force on the cycle due to the road is $200 \,N$ and is directly opposite to the motion. How much work does the road do on the cycle?

$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec{F} = (3\hat{i} + 4\hat{j})$ through two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

$A$ body is being raised to height $h$ from the surface of the Earth. What is the sign of work done by $(a)$ applied force and $(b)$ gravitational force?

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.

$A$ particle moves from $(1, 0, 3)$ to the point $(-3, 4, 5)$, when a force $F = \hat{i} + 5 \hat{k}$ acts on it. The amount of work done in joules is: