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Q.

A smooth track in the form of a quarter circle of radius 6m lies in the vertical plane. A particle moves from P1 to P2 under the action of forces F1F2 and F3. Force F1 is always toward P2 and is always 20 N in magnitude. Force F2 always acts horizontally and is always 30 N in magnitude. Force F3 always acts tangentially to the track and is of magnitude 15 N. Select the correct alternative(s)

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a

work done by F3 is 45π J

b

work done by F1 is 120 J

c

F1 is conservative in nature

d

work done by F2 is 180 J

answer is A, C, D.

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Detailed Solution

Work done by F1 is

 W1=P1P2F1cosθds

Where, ds=(6)d(2θ)=12 and F1=20N

W1=240π/40cosθdθW1=240sinπ4=1202J

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F1 is conservative because it is always directed towards a fixed point P2. Therefore, W1 can be directly calculated as

W1=F1P1P2=(20)(62)=1202J

Similarly, W2=F2OP2=(30)(6)=180 J

and W3=06(π/2)F3ds=03π15ds

       W3=15s03π=45π J

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