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Standard XI

Physics

Second Law of Motion

A monkey of 4...

Question

A monkey of 4 kg mass is climbing up on rope with v= $4 t^{2}$ , find tension in rope at t=5 sec.

40 N

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120 N

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160 N

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200 N

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Solution

The correct option is D 200 N
$v = 4 t^{2}$

$a = \frac{d v}{d t} = 8 t m / s^{2}$ (variable acc.)

at t=5 $s$ , a=40 $\frac{m}{s^{2}}$

for monkey

$f_{n e t} = T - m g = m a$

$T = m g + m a = m (g + a)$

$= 4 (10 + 40) = 200 N$

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Similar questions

Q. A monkey of 4 kg mass is climbing up on rope with v=

4 t^{2}

, find tension in rope at t=5 sec.

A monkey of mass

40

k g

climbs on a rope which can stand a maximum tension of

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N

. In which of the following cases will the rope break. When the monkey

(a) Climbs up with an acceleration of

6

m s^{- 2}

(b) Climbs down with an acceleration of

4

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5

m s^{- 1}

(d) Falls down the rope nearly freely under gravity.

(Ignore the mass of the rope)

Q. A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey (a) climbs up with an acceleration of 6 m s¯² (b) climbs down with an acceleration of 4 m s¯² (c) climbs up with a uniform speed of 5 m s¯¹ (d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

A monkey of mass 20kg climbs on a rope which can stand a maximum tension of 300 N.in which of the following cases the rope breaks? The monkey

Options

Climbs up with an acceleration of 6m/s2

Climbs up with a uniform speed of 5m/s

Climbs down with an acceleration of 4m/s2

Falls down the rope nearly freely under gravity

Ans :1st option

Q. Three monkeys A,B,C with masses of 10,15,8 kg respectively are climbing up and down the rope suspended from

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. At the instant represented,

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is descending the rope with an acceleration of

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and

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is pulling himself up with an acceleration of

1.5 m / s^{2}

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A monkey of 4 kg mass is climbing up on rope with v=4t2, find tension in rope at t=5 sec.

The correct option is D 200 Nv=4t2 a=dvdt=8t m/s2 (variable acc.) at t=5 s, a=40 m/s2 for monkey fnet=T−mg=ma T=mg+ma=m(g+a) =4(10+40)=200 N

A monkey of 4 kg mass is climbing up on rope with v= $4 t^{2}$ , find tension in rope at t=5 sec.

The correct option is D 200 N
$v = 4 t^{2}$

$a = \frac{d v}{d t} = 8 t m / s^{2}$ (variable acc.)

at t=5 $s$ , a=40 $\frac{m}{s^{2}}$

for monkey

$f_{n e t} = T - m g = m a$

$T = m g + m a = m (g + a)$

$= 4 (10 + 40) = 200 N$