Work Done by a Force

The area of the acceleration-displacement curve of a body gives

Find the work done by the gravity on the man of mass $60 \mathrm{kg}$ while going up the escalator of height $4 \mathrm{m}$.(Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

Johnny and his sister Jane race up a hill. Johnny weighs twice as much as Jane and takes twice the time as long as Jane to reach the top. Compared to Jane:

A block of mass $50 \mathrm{kg}$ just slides over a distance of $1 \mathrm{m}$. If the coefficient of friction between their surfaces is $0.2$, the work done against friction is

A cubical vessel of height $1 \mathrm{m}$ is full of water. What is the amount of work done in pumping water out of the vessel?

[Take $g=10 {\mathrm{ms}}^{-2}$]

When a force is applied on a moving body, its motion is retarded. Then the work done is

A body of mass $3 \mathrm{kg}$ is under a force, which causes a displacement in the body given by $s=$ $\frac{t^3}{3}$ (in meters). Find the work done by the force in the first $2$ seconds.

A block of mass $50 \mathrm{kg}$ just slides over a distance of $1 \mathrm{m}$. If the coefficient of friction between their surfaces is $0.2$, the work done against friction is

A cubical vessel of height $1 \mathrm{m}$ is full of water. What is the amount of work done in pumping water out of the vessel?

[Take $g=10 {\mathrm{ms}}^{-2}$]

When a force is applied on a moving body, its motion is retarded. Then the work done is

Force of $4 \mathrm{N}$ is applied on a body of mass $20 \mathrm{kg}$. Find the work done in Joules in $3$rd second.

A $200 \mathrm{kg}$ cart is pushed slowly up the incline. The work done by the pushing force in order to move the cart up along the incline on to the platform $1.5 m$ above the ground is (neglecting friction, $\mathrm{g}=9.8 \mathrm{m}/{\mathrm{s}}^2$)

A body, possessing kinetic energy $T\mathit{,}$ moving on a rough horizontal surface, is stopped in a distance $y.$ The frictional force exerted on the body is :-

The system of two masses $2 \mathrm{kg}$ and $3 \mathrm{kg}$ shown in the figure is released from rest. The work done on $3 \mathrm{kg}$ block by the force of gravity during first $2 \mathrm{seconds}$ of its motion is $\left(g\mathit{=}10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

The force acting on the block is $10\sqrt{2} \mathrm{N}$ with an angle $135°.$ Calculate the work done when the displacement is $\mathrm{\Delta x}=5 \mathrm{m}$

A force $\overrightarrow{F}=\left(3x^2\mathrm{N}\right)\hat{\mathrm{i}}+\left(4 \mathrm{N}\right)\hat{\mathrm{j}}$, (where $x$ is in meters), acts on a particle. Assuming that only kinetic energy is changed the net work done on the particle as it moves from position $\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}\right) \mathrm{m}$ to $\left(3\hat{\mathrm{i}}+1\hat{\mathrm{j}}\right) \mathrm{m}$

Consider a drop of rainwater having mass $1 \mathrm{g}$ falling from a height of $1 \mathrm{km}$. It hits the ground with a speed of $50 \mathrm{m} {\mathrm{s}}^{-1}$. Take $\mathrm{g}$ constant with a value $10 \mathrm{m} {\mathrm{s}}^{-2}$. The work by the
$\mathrm{i})$ Gravitational force and
$\mathrm{ii})$ Resistive force of air is

Force acting on a particle moving in a straight line varies with the velocity of the particle as $F=\frac{K}{v}$. Here, $K$ is a constant. The work done by this force in time $t$ is