(2) If a person is pushing a box inside a moving train, the work done in the frame of train will \[\overrightarrow{F}.\overrightarrow{s\,}\] while in the frame of earth will be \[\overrightarrow{F}.\,(\overrightarrow{s\,}+{{\overrightarrow{\,s}}_{0}})\] where \[{{\overrightarrow{\,s}}_{0}}\] is the displacement of the train relative to the ground. 
(2) In conservative field work done by the force (line integral of the force i.e. \[\int{\overrightarrow{F}.d\overrightarrow{l\,}}\]) over a closed path/loop is zero.
\[{{W}_{A\to B}}+{{W}_{B\to A}}=0\]
or \[\oint{\vec{F}.d\vec{l}=0}\]
Conservative force : The forces of these type of fields are known as conservative forces.
Example : Electrostatic forces, gravitational forces, elastic forces, magnetic forces etc and all the central forces are conservative in nature.
If a body of mass m lifted to height h from the ground level by different path as shown in the figure
Work done through different paths
\[{{W}_{I}}=F.\,s=mg\times h=mgh\]
\[{{W}_{II}}=F.\,s=mg\sin \theta \times l=mg\sin \theta \times \frac{h}{\sin \theta }=mgh\]
\[{{W}_{III}}=mg{{h}_{1}}+0+mg{{h}_{2}}+0+mg{{h}_{3}}+0+mg{{h}_{4}}\]
\[=mg({{h}_{1}}+{{h}_{2}}+{{h}_{3}}+{{h}_{4}})=mgh\]
\[{{W}_{IV}}=\int{\overrightarrow{F}.\,d\overrightarrow{s\,}}=mgh\]
It is clear that \[{{W}_{I}}={{W}_{II}}={{W}_{III}}={{W}_{IV}}=mgh\].
Further if the body is brought back to its initial position A, similar amount of work (energy) is released from the system, it means \[{{W}_{AB}}=mgh\] and \[{{W}_{BA}}=-mgh\].
Hence the net work done against gravity over a round trip is zero.
\[{{W}_{Net}}={{W}_{AB}}+{{W}_{BA}}\]\[=mgh+(-mgh)=0\]
i.e. the gravitational force is conservative in nature.
Non-conservative forces : A force is said to be non-conservative if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions and for complete cycle this work done can never be zero.
Example: Frictional force, Viscous force, Airdrag etc.
If a body is moved from position A to another position B on a rough table, work done against frictional force shall depend on the length of the path between A and B and not only on the position A and B.
\[{{W}_{AB}}=\mu mgs\]
Further if the body is brought back to its initial position A, work has to be done against the frictional force, which opposes the motion. Hence the net work done against the friction over a round trip is not zero.
\[{{W}_{BA}}=\mu mgs.\]
\[\therefore \,{{W}_{Net}}={{W}_{AB}}+{{W}_{BA}}=\mu mgs+\mu mgs=2\mu mgs\ne 0.\]
i.e. the friction is a non-conservative force. 
Let F be the average value of variable force within the interval dx from position x to (x + dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in displacing it from position \[{{x}_{i}}\] to \[{{x}_{f}}\] will be equal to the sum of areas of all the such strips
\[dW=\overrightarrow{F}\,dx\]
\[\therefore \,W=\int_{{{x}_{i}}}^{{{x}_{f}}}{dW=\int_{{{x}_{i}}}^{{{x}_{f}}}{F\,dx}}\]
\[\therefore \,W=\int_{{{x}_{i}}}^{{{x}_{f}}}{(\text{Area}\,\text{of}\,\text{strip}\,\text{of}\,\text{width}\,dx)}\]
\[\therefore \,W=\text{Area}\,\text{under}\,\text{curve}\,\text{between}\,{{x}_{i}}\,\text{and}\,{{x}_{f}}\]
i.e. Area under force-displacement curve with proper algebraic sign represents work done by the force. 
The total work done in going from A to B as shown in the figure is
\[W=\int_{A}^{B}{\,\overrightarrow{F}.\,d\overrightarrow{s\,}=\int_{A}^{B}{\,(F\cos \theta )ds}}\]
In terms of rectangular component \[\overrightarrow{F}={{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}+{{F}_{z}}\hat{k}\]
\[d\overrightarrow{s\,}=dx\hat{i}+dy\hat{j}+dz\hat{k}\]
\[\therefore \,\,\,W=\int_{A}^{B}{({{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}+{{F}_{z}}\hat{k})}.(dx\hat{i}+dy\hat{j}+dz\hat{k})\]
or \[W=\int_{{{x}_{A}}}^{{{x}_{B}}}{{{F}_{x}}dx+\int_{{{y}_{A}}}^{{{y}_{B}}}{{{F}_{y}}dy}}+\int_{{{z}_{A}}}^{{{z}_{B}}}{{{F}_{z}}dz}\] | Absolute units | Gravitational units |
| Joule [S.I.]: Work done is said to be one Joule, when 1 Newton force displaces the body through 1 metre in its own direction. From, W = F.s 1 Joule = 1 Newton x 1 m | kg-m [S.I.]: 1 kg-m of work is done when a force of 1kg-wt. displaces the body through 1m in its own direction. From W = F s 1 kg-m = 1 kg-wt x 1 m = 9.81 N x 1 metre = 9.81 Joule |
| erg [C.G.S.] : Work done is said to be one erg when 1 dyne force displaces the body through 1 cm in its own direction. From W = F s \[1\,erg=1dyne\times 1cm\] Relation between Joule and erg 1 Joule = 1 N x 1 m = \[{{10}^{5}}\] dyne x \[{{10}^{2}}\] cm = \[{{10}^{7}}\] dyne x cm = \[{{10}^{7}}\] erg | gm-cm [C.G.S.] : 1 gm-cm of work is done when a force of 1gm-wt displaces the body through 1cm in its own direction. From W = F s 1 gm-cm = 1gm-wt x 1cm. = 981 dyne x 1cm = 981 erg |
\[{{0}^{o}}\le \theta <{{90}^{o}}\]
The positive work signifies that the external force favours the motion of the body.
Example: (i) When a person lifts a body from the ground, the work done by the (upward) lifting force is positive
(ii) When a lawn roller is pulled by applying a force along the handle at an acute angle, work done by the applied force is positive.
(iii) When a spring is stretched, work done by the external (stretching) force is positive.
Maximum work : \[{{W}_{\max }}=F\,s\]
When \[\cos \theta =\text{maximum}=1\] i.e. \[\theta ={{0}^{o}}\]
It means force does maximum work when angle between force and displacement is zero.
Negative work
Negative work means that force (or its component) is opposite to displacement i.e.
\[{{90}^{o}}<\theta \le {{180}^{o}}\]
The negative work signifies that the external force opposes the motion of the body.
Example: (i) When a person lifts a body from the ground, the work done by the (downward) force of gravity is negative.
(ii) When a body is made to slide over a rough surface, the work done by the frictional force is negative.
Minimum work : \[{{W}_{\min }}=-F\,s\]
When \[\cos \theta =\text{minimum}=-1\] i.e \[\theta ={{180}^{o}}\]
It means force does minimum [maximum negative] work when angle between force and displacement is \[{{180}^{o}}\].
(iii) When a positive charge is moved towards another positive charge. The work done by electrostatic force between them is negative.
| Zero work | |
| Under three condition, work done becomes zero \[W=Fs\cos \theta =0\] | |
| (1) If the force is perpendicular to the displacement \[\mathbf{[}\overrightarrow{F}\,\bot \,\overrightarrow{s\,}\mathbf{]}\] Example: (i) When a coolie travels on a horizontal platform with a load on his head, work done against gravity by the coolie is zero. (ii) When a body moves in a circle the work done by the centripetal force is always zero. (iii) In case of motion of a charged particle in a magnetic field as force \[[\overrightarrow{F}=q(\overrightarrow{v\,}\times \overrightarrow{B})]\] is always perpendicular to motion, work done by this force is always zero. |  |
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(2) If there is no displacement more...
Let a constant force \[\overrightarrow{F}\] be applied on the body such that it makes an angle q with the horizontal and body is displaced through a distance s.
By resolving force \[\overrightarrow{F}\] into two components :
(i) F cos \[\theta \] in the direction of displacement of the body.
(ii) F sin \[\theta \] in the perpendicular direction of displacement of the body.
Since body is being displaced in the direction of \[F\cos \theta \], therefore work done by the force in displacing the body through a distance s is given by
\[W=(F\cos \theta )\,s=Fs\cos \theta \] or \[W=\overrightarrow{F}.\overrightarrow{s\,}\]
Thus work done by a force is equal to the scalar (or dot product) of the force and the displacement of the body.
If a number of forces \[{{\overrightarrow{F}}_{1}},\,{{\overrightarrow{F}}_{2}},\,{{\overrightarrow{F}}_{3}}......{{\overrightarrow{F}}_{n}}\] are acting on a body and it shifts from position vector \[{{\overrightarrow{\,r}}_{1}}\] to position vector \[{{\overrightarrow{\,r\,}}_{2}}\] then \[W=({{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}+....{{\overrightarrow{F}}_{n}})\,.({{\overrightarrow{\,r}}_{2}}-{{\overrightarrow{\,r}}_{1}})\]
The terms 'work', 'energy' and 'power' are frequently used in everyday language. A farmer clearing weeds in his field is said to be working hard. A woman carrying water from a well to her house is said to be working. In a drought affected region she may be required to carry it over large distances. If she can do so, she is said to have a large stamina or energy. Energy is thus the capacity to do work. The term power is usually associated with speed. In karate, a powerful punch is one delivered at great speed. In physics we shall define these terms very precisely. We shall find that there is a loose correlation between the physical definitions and the physiological pictures these terms generate in our minds.
Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.
When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left.
This force (ma) is action force by a block on cart.
Now block will remain static w.r.t. cart. If friction force \[\mu R\ge mg\]
\[\Rightarrow \] \[\mu ma\ge mg\] \[[\text{As}\,R=ma]\]
\[\Rightarrow \] \[a\ge \frac{g}{\mu }\]
\[\therefore \] \[{{a}_{\min }}=\frac{g}{\mu }\]
This is the minimum acceleration of the cart so that block does not fall.
and the minimum force to hold the block together
\[{{F}_{\min }}=(M+m)\,{{a}_{\min }}\]
\[{{F}_{\min }}=(M+m)\,\frac{g}{\mu }\]
A body of mass m which is placed at the top of the wedge (of height h) starts moving downward on a rough inclined plane.
Loss of energy due to friction = FL (Work against friction)
PE at point A = mgh
KE at point B = \[\frac{1}{2}m{{u}^{2}}\]
By the law of conservation of energy
i.e. \[\frac{1}{2}m{{v}^{2}}=mgh-FL\]
\[v=\sqrt{\frac{2}{m}(mgh-FL)}\] Current Affairs CategoriesArchive
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