$A$ force $F = (2 \hat{i} + 4 \hat{j}) \text{ N}$ is applied on an object of mass $M$. What is the work done by this force in moving the object horizontally along the $X$-axis by $3 \text{ m}$ (in $\text{ J}$)?

$A$ 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.....$^o$

$A$ horizontal force of $10 \ N$ is applied on a block of mass $1.5 \ kg$ which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of $6 \ s$ from the beginning of the motion is (Acceleration due to gravity $= 10 \ m/s^2$; the coefficient of kinetic friction between the block and the surface is $0.2$). (in $J$)

At time $t=0$,a particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time $t$,the net force acting on it must be proportional to

$A$ force $\vec F = 2\hat i - 3\hat j + 7\hat k \text{ (N)}$ acts on a particle which undergoes a displacement of $\vec S = 7\hat i + 3\hat j - 2\hat k \text{ (m)}$. Calculate the work done by the force in $J$.

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 $\text{ J}$)