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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:

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Explanation:

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R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

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Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

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