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NEET PHYSICSMedium

With what minimum acceleration can a fireman slide down a rope while the breaking strength of the rope is 23\frac{2}{3} of his weight?

A

23g\frac{2}{3}g

B

gg

C

13g\frac{1}{3}g

D

Zero

Step-by-Step Solution

  1. Analyze Forces: The forces acting on the fireman descending the rope are:
  • Weight (W=mgW = mg) acting vertically downwards.
  • Tension (TT) acting vertically upwards (provided by the rope).
  1. Equation of Motion: According to Newton's Second Law, the net force downward produces the acceleration (aa): mg−T=mamg - T = ma T=m(g−a)T = m(g - a)
  2. Constraint: The rope breaks if the tension exceeds its breaking strength. The maximum tension the rope can withstand is given as 23\frac{2}{3} of the weight. Tmax=23mgT_{max} = \frac{2}{3}mg
  3. Minimum Acceleration: To slide down safely, the tension must be less than or equal to TmaxT_{max}. A lower acceleration requires a higher tension (holding tighter). Therefore, the minimum acceleration corresponds to the maximum possible tension (just on the verge of breaking). m(g−amin)=Tmaxm(g - a_{min}) = T_{max} m(g−amin)=23mgm(g - a_{min}) = \frac{2}{3}mg
  4. Calculation: Cancel mass mm and solve for amina_{min}: g−amin=23gg - a_{min} = \frac{2}{3}g amin=g−23ga_{min} = g - \frac{2}{3}g amin=13ga_{min} = \frac{1}{3}g
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