Question

two block of mass 10 kg and 2kg respectively are connected by an ideal string passing over a fixed smooth Pulley as shown in figure a monkey of 8 kg started climbing the string with constant acceleration 2metre per second square with respect to string t=0 initially the monkey is 2.4 m from the pulley find time taken by the monkey to reach

Answer 1

1 second option first

Answer 2

Answer:

Explanation:

Concept:

Acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates.

Given:

Two block of mass 10 kg and 2kg respectively are connected by an ideal string passing over a fixed smooth Pulley.

A monkey of 8 kg started climbing the string with constant acceleration 2metre per second square with respect to string t=0 initially the monkey is 2.4 m from the pulley.

Find:

Find time taken by the monkey to reach

Solution:

Given,

mass of the monkey, $m_{m}$ = 8 kg

and its acceleration w.r.t the rope, $a_{o}$ = 2 m $s^{-2}$

If the acceleration of rope is a downward, acceleration of monkey w.r.t ground, $a_{m}$ = $a_{o}$ - $a$

From free body diagram of monkey:

$T-8 g-T_{1}=8 \times a_{m}=8(2-a) \quad \ldots(1)$

From free body diagram of block of mass $2 \mathrm{~kg}$ :

$2 \mathrm{~g}-\mathrm{T}_{1}=2 \mathrm{a}$   ------(2)

From free body diagram of block of mass $10 \mathrm{~kg}$:

$\mathrm{T}-10 \mathrm{~g}=10 \mathrm{a}$   ------(3)

From (1), (2) and (3), we have

  $(10 a+10 g)-8 g-(2 g-2 a)=16-8 a \Rightarrow \mathrm{a}=\frac{4}{5}$

Now, net acceleration of monkey,

$\begin{aligned}&a_{m}=a_{0}-a=2-\frac{4}{5} \\&\Rightarrow a_{m}=1.2 \mathrm{~ms}^{-2}\end{aligned}$

Using equation of motion, time taken by the monkey to reach the pulley,

$\mathrm{t}=\sqrt{\frac{2 \times \mathrm{S}}{\mathrm{a}_{\mathrm{m}}}}=\sqrt{\frac{2 \times 2.4}{1.2}}=2 \mathrm{sec}$

Hence option (2) is correct.

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