The sign of work done by a force on a body is importantto understand.State carefully if the following quantities are positiveor negative : (a) work doneby a manin liftinga bucketout ofa wellby meansof a ropetiedto the bucket. (b) work done by gravitationalforce in the abovecase , (c ) work doneby friction on a bodysidingdown an inclined plane , (d ) workdone byan applied force on a bodymovingon a roughhorizontalplane wothuniform velocity . (d) work doneby the resistiveforceof air on a vibrating pendulum in bringingit to rest .

The sign of work done by a force on a body is importantto understand.S

1.	The sign of work done by a force on a body is importantto understand.State carefully if the following quantities are positiveor negative : (a) work doneby a manin liftinga bucketout ofa wellby meansof a ropetiedto the bucket. (b) work done by gravitationalforce in the abovecase , (c ) work doneby friction on a bodysidingdown an inclined plane , (d ) workdone byan applied force on a bodymovingon a roughhorizontalplane wothuniform velocity . (d) work doneby the resistiveforceof air on a vibrating pendulum in bringingit to rest .
Answer» Solution :We know that work`W = Fd cos theta` , where `theta ` is the angle between `vec(F ) and vec(d)` . (a) Here , DISPLACEMENT of a BUCKET is in the directionof force exterted on bucket so `theta = 0^(@)` ` :. Cos 0^(@) =1 "" : W = Fd cos 0^(@) =Fd ` ` :. ` Work is positive . (b) Work done is negative ,because gravitational force is in the downward and displacementof bucketis upwarddirection so , `theta = 180^(@)` ` :.W = Fd cos 180^(@)` ` :. W = - Fd "" [ :. cos 180^(@) =-1]` (c ) Work doneis negative , because force offriction is acting against the displacemnt so `theta = 180^(@)` ` :. W = fd cos 180^(@)` ` :.W = Fd ` (e) Work doneis negative , because the RESISTIVE force of air OPPOSES the motion so , `theta = 180^(@)` ` :. W - Fd cos 180^(@)` ` :. W = - Fd `