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nirvana33 [79]
3 years ago
7

Two person A and B of weight 60kg and 40kg respectively stand facing each other, and pull on a light rope streched between them.

Find the ratio of distance covered by them when they meet if their intial separation was 5 metre
Physics
1 answer:
goblinko [34]3 years ago
6 0

To be able to successfully pull the rope with balance, the work done by person A should be just equal to the work done by person B.

<span>workA = workB  ----> 1</span>

Where: work = force * distance

In this case, force = mass * gravity

So work = mass * gravity * distance

Rewriting equation 1:

<span>massA * gravity * distanceA = massB * gravity * distanceB  ---->2</span>

We know that distanceA + distanceB = 5 m, therefore distanceA = 5 – distanceB. Substituting this to equation 2 and cancelling constant g:

massA * distanceA = massB * distanceB

60 kg * (5 – distanceB) = 40 kg * distanceB

distanceB = 3 m

distanceA = 5 – distanceB = 2 m

 

<span>ratio is: distanceA / distanceB = 2/3 = 0.67</span>

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Answer:

\mu =0.75

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<u>Frictional Force </u>

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The centripetal acceleration a_c is computed as

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Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

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\boxed{\mu =0.75}

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