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Newton's third law of motion

In order to raise a block of mass 100 kg a man of mass 60 kg f

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Medium

Solving time: 3 mins

In order to raise a block of mass $100$ kg a man of mass $60$ kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $\frac{5 g}{4}$ relative to rope. The tension in the rope is $(g = 10 m s^{- 2})$ .

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Updated on: Apr 10, 2023

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Liam Discussed

In order to raise a block of mass

100

kg a man of mass

60

kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration

\frac{5 g}{4}

relative to rope. The tension in the rope is

(g = 10 m s^{- 2})

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

T - 100 g = 100 a

T - 60 g = 60 (\frac{5}{4} g - a)

- 40 g = 160 a - 75 g

160 a = 35 g

a = \frac{35}{160} \times 10 = \frac{35}{16} m / s^{2}

T = 100 g + 100 a

1000 + 100 \times \frac{35}{16}

\frac{4000 + 875}{4} = \frac{4875}{4} = 1218.75 N

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Question Text	In order to raise a block of mass $100$ kg a man of mass $60$ kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $\frac{5 g}{4}$ relative to rope. The tension in the rope is $(g = 10 m s^{- 2})$ .
Updated On	Apr 10, 2023
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 14
Upvotes	1413
Avg. Video Duration	6 min

In order to raise a block of mass $100$ kg a man of mass $60$ kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $\frac{5 g}{4}$ relative to rope. The tension in the rope is $(g = 10 m s^{- 2})$ .

Text solutionVerified

Filo tutor solutions (14)

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In order to raise a block of mass $100$ kg a man of mass $60$ kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $\frac{5 g}{4}$ relative to rope. The tension in the rope is $(g = 10 m s^{- 2})$ .