Physics, asked by ramojujayasri, 11 months ago

The minimum acceleration with which a person can slide down a rope whose breaking strength is 60percent of his weight

Answers

Answered by wwwausalishivani
20

Answer: 2g/5

Explanation:

T = mg - ma

T= breaking force= 60×mg/100

m(g-a) = 60×mg /100

a= 10g - 6g /10

a=4g/10

a= 2g/5

Answered by ansiyamundol2
2

Answer:

The minimum acceleration is \frac{2}{5}g.

Explanation:

Given that :

Inertia force (ma)= minimum

Force due to gravity (mg)=\frac{3}{5}

T-mg+ma=0\\ma=mg-T

ma=mg-\frac{3}{5}mg

Dividing throughout with m : a=g-\frac{3}{5}g\\

a=\frac{5g-3g}{5}\\ a=\frac{2g}{5}

D'Alembert's concept is used in this procedure.

  • We can solve this utilising dynamic equilibrium if we add the inertia force to the rope. Because he is accelerating downward, the inertia force acts upward F = ma (opposite to direction of acceleration). The human is still under the influence of tension and gravity.
  • If a=0, the rope tension is equal to the person's weight.
  • NOTE: If he accelerates up the rope, the inertia force acts downward, causing the rope tension to exceed the person's weight.

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