Work Done by a Variable Force

A force $\overrightarrow{F}=\left(2+3x\right)\hat{i}$ acts on a particle in the $x$ direction where $F$ is in Newton and $x$ is in meter. The work done by this force during a displacement from $x = 0$ to $x=4 \mathrm{m}$ is _____ $\mathrm{J}$.

A block of mass $10 \mathrm{kg}$ is moving along $x$-axis under the action of force $F=5x \mathrm{N}$. The work done by the force in moving the block from $x=2 \mathrm{m}$ to $4 \mathrm{m}$ will be _______ $\mathrm{J}$.

A variable force $F=5Kx \mathrm{N}$ acts on a body moving along x-axis. Find the work done by this force in displacing the body from $x= 2 \mathrm{m}$ to $x=5 \mathrm{m}$ ($K$ is a constant).

Force acting on a particle moving along $x-$axis is given by $\overrightarrow{F}=(2+3x)\hat{\mathrm{i}}$. The work done by this force from $x=0$ to $x=4 \mathrm{m}$ is

A particle starts to move along the path which locus is given as $x^2=\left(8y\right)$ $(x\&y \text{are in} \mathrm{m})$, due to applied force $\overrightarrow{F}=(ax\hat{\mathrm{i}}+by\hat{\mathrm{j}} ) \mathrm{N}$ $(a>0 \text{and} b>0 \text{and} x \& y \text{are in metre})$. The amount of work done in displacing particle from $x=0$ to $x=1$ metre is

A force,$F=(10+x) \mathrm{N}$ acts on a particle of mass $4 \mathrm{kg}$ in $x$ direction, where $x$ is in meter. Find the work done by the force during the time interval its acceleration changes from$2.5 \mathrm{m} {\mathrm{s}}^{-2}$ to  $5.0 \mathrm{m} {\mathrm{s}}^{-2}$
 

A particle of mass $0.5 \mathrm{kg}$is subjected to a force which varies with distance as shown in graph

If the speed of the particle at $x= 0$is $4 \mathrm{m} {\mathrm{s}}^{-1}$, then its speed at $x=8 \mathrm{m}$ is

If the potential energy at point $\mathrm{O} (0, 0)$ is ${\mathrm{U}}_0= 10 \mathrm{J}$, find the potential energy at point $\mathrm{D} (4, 0)$ corresponding to the given graph.

Calculate the net work done by the force $F$ whose variation with position has been shown as above.

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

A block of mass is placed at origin $(0, 0)$ and a force, $\mathrm{F}=(2\mathrm{x} + 2) \mathrm{N}$, is applied on the block that it reaches at some other point. $\mathrm{P} (5, 0)$. Find the work done by the force.
 

A block of mass $m$ is placed at origin $\left(0,0\right)$ and a force, $F=\left(2+2x\right) \mathrm{N}$, is applied on the block so that it reaches at some other point, $P \left(5,0\right)$. Find the work done by the force.

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Calculate the net work done by the force $F$ whose variation with position has been shown as above.

A force $F=k\left[y\hat{\mathrm{i}}+x\hat{\mathrm{j}}\right] \mathrm{N}$ where $k$ is a positive constant act on a particle moving in $x-y$ plane starting from the point $\left(3,5\right)$, the particle is taken along a straight line to $\left(5,7\right)$. The work done by the force is:

A particle is moving in anticlockwise direction in a circle under the influence of a force,  $\overrightarrow{F}= \left[\left(x^2-y\right)\hat{\mathrm{i}}+\left(x+5y-z\right)\hat{\mathrm{j}}+\left(3x-4y+z\right)\hat{\mathrm{k}}\right] \mathrm{N}$, where $x, y$ and $z$ are in meter. The circle lies on the $xy-$plane with its center at the origin and has a radius of $2 \mathrm{m}$. The work done by the force as the particle completes one revolution is (in Joule)[Take $\pi = 3.14$]

Force acting on a particle varies with displacement as shown in figure. The workdone by this force on the particle from $x=-8 \mathrm{m}$ to $x=+8 \mathrm{m}$

A small block of mass $m$ is moving under the effect of a force which is given by $\overrightarrow{F}=\left(3x\hat{j}+3y\hat{i}\right)\mathrm{N}$. If this force displaces the block from point $A\left(2 \mathrm{m},3 \mathrm{m}\right)$ to another point $B\left(1 \mathrm{m},4 \mathrm{m}\right)$, then work done by the force will be

While calculating work done by a variable force, why do we calculate the work done for small displacements $∆x$?

If the force on a body is varying, then how will you find the work done by the force?

A position-dependent force $F=7-2x+3x^2$ newton acts on a small body of mass $2 \mathrm{kg}$ and displaces it from $x=0$ to $x=5$. Calculate the work done.