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Question

Physics Question on laws of motion

With what maximum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?

A

g3\frac{g}{3}

B

23g\frac{2}{3} g

C

32g\frac{3}{2}g

D

g2\frac{g}{2}

Answer

g3\frac{g}{3}

Explanation

Solution

Net downward force on fireman is
ma=mgTma = mg - T
As Tmin=23mgT_{min} = \frac{2}{3} mg
mamax=mg23mg\therefore \, ma_{max} = mg - \frac{2}{3} mg
or amax=g3a_{max} = \frac{g}{3}