In order to raise a mass of 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 acceleration 5g/4 relative to the rope.The tension in the rope is (take $g = 10 m s^{2}$ )

\frac{4875}{8} N

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\frac{4875}{2} N

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\frac{4875}{4} N

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\frac{4875}{6} N

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Solution

The correct option is C $\frac{4875}{4} N$
Let $a_{r}$ be the acceleration of man relative to a and $a_{m}$ be the acceleration of man.
$T - m g = m a_{m}$
$T - M g = M a$
Now $a_{m} = a_{r} - a$
$\Rightarrow g (M - m) = M a - m a_{r} + m_{a}$
$\Rightarrow a = \frac{[g (M - m) + m a_{r}]}{(M + m)}$
Given $M = 100 k g m = 60 k g$ $a_{r} = \frac{5}{4} g$
$\Rightarrow a = \frac{[(- 10) \times 40 + 750]}{160}$
$a = \frac{35}{16}$
$T = m (g + a) = \frac{4875}{4} N$

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Similar questions

In order to raise a mass of 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 acceleration 5 $\frac{g}{4}$ relative to the rope. The tension in the rope is (take g = 10 $m s^{- 2}$ )

Q. in order to raise a mass of 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 5 g/ 4

Q. Block A of mass

m / 2

is connected to one end of light rope which passes over a pulley as shown in the figure. Man of mass

m

climbs the other end of rope with a relative acceleration of

g / 6

with respect to rope. Find acceleration of block A and tension in the rope.

Q. Two men of masses

m_{1} a n d m_{2}

hold on the opposite ends of a rope passing over a frictionless pulley. The mass

m_{1}

climbs up the rope with an acceleration of

1.2 m / s^{2}

relative to the rope. The man

m_{2}

climbs up the rope with an acceleration of

2.0 m / s^{2}

relative to the rope. They start from rest and are initially separated by

5 m

. (Given

m_{1} = 40 k g

and

m_{2} = 60 k g

)

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In order to raise a mass of 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 acceleration 5g/4 relative to the rope.The tension in the rope is (take g=10ms2)

The correct option is C 48754NLet ar be the acceleration of man relative to a and am be the acceleration of man.T−mg=mamT−Mg=MaNow am=ar−a⇒g(M−m)=Ma−mar+ma⇒a=[g(M−m)+mar](M+m)Given M=100kgm=60kg ar=54g⇒a=[(−10)×40+750]160a=3516T=m(g+a)=48754N

In order to raise a mass of 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 acceleration 5g/4 relative to the rope.The tension in the rope is (take $g = 10 m s^{2}$ )