Question

A fixed wedge ABC of mass m is in the shape of an equi triangle. Initially a achain of length 2l and mass m rests on the wedge. The chain is being pulled slowly by a force F. Work done by gravity till the time the chain leaves the wedge?

Answer 1

Explanation:

Initial center of mass of the chain is at a height of L/2 sin60= √3/4 L

Final center of mass is at a depth L

total distance moved by center of mass of the chain= (√3/4 +1)L

Work done by the gravitational force= mgL (√3+4)/4

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Answer 2

Answer:

Hey mate here is your answer.....

Dear student

Side of the equilateral triangle

=

l

=l

Initially centre of mass of the chain

=

L

2

sin

60

=

√

3

4

L

=L2sin⁡60=34L

Final centre of mass is at depth L

Total distance moved by the centre of mass of chain

=

√

3

4

L

+

L

=34L+L

=

L

(

√

3

4

+

1

)

=L(34+1)

Work done

W

=

F

.

l

W=F.l

=

m

g

l

=mgl

=

L

(

√

3

4

+

1

)

m

g

=L(34+1)mg

=

m

g

l

(

√

3

4

+

1

)

=mgl(34+1)

hope it's helpful for you...

thanks