Question

CONSIDER THE MASS AS 5M AS WELL AS M (TWO CASES).PLEASE DO ANSWER WITH EXPLANATION QUICKLY ​

Answer 1

Solution:-

Let AN = x = NB

ON = z

When we pull the both string

$\rm \: \frac{dz}{dt} = v_m \: \: \: \: \: \: \: \: (v_m = velocity \: of \: mass)</p><p>$

So l is also change so we write

$\rm \: \frac{dl}{dt} = u$

In triangle OAN

Using phythogoeros theorem

l² = x² + z²

l and z is constantly change with time

so x is fixed

differentiate w . r .t 't'

$\rm \: l {}^{2} \: \frac{dl}{dt} = {x}^{2} + \frac{dz}{dt} \: {z}^{2}$

So x is constant

$\rm \: 2l \: \: \: \frac{dl}{dt} = 0 + 2z \: \: \frac{dz}{dt}$

Given

$\rm \: \frac{dl}{dt} = u$

$\rm \: \frac{dz}{dt} = v_m$

We get

$\rm \: lu = zv_m$

$\rm \: v_m = \frac{lu}{z}$

$\rm \: \rm \: v_m = \frac{u}{ \frac{z}{l} }$

$\rm \: v_m = \frac{u}{ \cos( \theta) }$

$\rm \: \theta = 60 \degree \\ \cos(60) \degree = \frac{1}{2}$

$\rm \: v_m = \frac{u}{ \frac{1}{2} }$

$\rm \: v_m = 2u$

option c is correct

Answer 2

Answer:

Let AN = x = NB

Let AN = x = NBON = z

Let AN = x = NBON = zWhen we pull the both string

Let AN = x = NB

Let AN = x = NBON = z

Let AN = x = NBON = zWhen we pull the both string

Let AN = x = NBON = zWhen we pull the both string\rm \: \frac{dz}{dt} = v_m

(v_m = velocity of mass) dtdz =v m

(v m =velocityofmass)So l is also change so we write\rm {dl}/{dt}= u

dtdl =uIn triangle OAN Using

phythogoeros theoreml² = x² + z²l and z is constantly change with timeso x is fixeddifferentiate w . r .t 't'[/tex]

c