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A 3.0kg block has a speed of 2m//s A and...

A 3.0kg block has a speed of `2m//s` A and `6m//s` at B. If the distance from A and B along the curve is 12m, how large a frictional force acts on it ? Assuming the same friction, how far from B will it stop ?

Text Solution

Verified by Experts

The correct Answer is:
24.5
Change in total energy of block = Work done against friction
`rArr mg(h- x)+ (1)/(2) m(v_(A)^(2)- v_(B)^(2))= f_(s)`
`rArr 3 xx 9.8 xx (4-1) + (1)/(2) xx 3 xx (2^(2) -6^(2)) = f xx 12`
`rArr 9 xx 9.8 - 3 xx 16 = f xx 12`
or `f= (40.2)/(12)= 3.35N`
Total energy at `B= 3 xx 9.8 xx 1+ (1)/(2) xx 3 xx 6^(2)`
`rArr 29.4 + 54= 83.4`
This energy is used in doing work against friction
`therefore f xx s.= 83.4`
or `s.= (83.4)/(f)= (83.4)/(3.35)= 24.5m`
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