Question

in the figure the ends P and Q of an unstrechable string move downwards with uniform speed v. mass M moves upward with what speed?

Answer 1

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Answer 2

Given that,

In the figure the ends P and Q of an un stretchable string move downwards with uniform speed v.

Mass M moves upward.

According to figure ,

Let the length of the string from A to C is l.

Suppose, The distance between A and N is x and x does not change so it is constant.

The distance between N and C is y.

When we pull the P and Q with uniform speed v then y will be change.

According to figure,

The change in height will be equal to the velocity of mass.

$\dfrac{dy}{dt}=v_{M}$

The change in length of the string will be equal to the velocity of P and Q.

$\dfrac{dl}{dt}=v$

We need to calculate the velocity of mass

Using diagram,

In triangle CAN,

$l^2=x^2+y^2$

Where, x = constant

On differentiating with respect to t

$2l\dfrac{dl}{dt}=0+2y\dfrac{dy}{dt}$

Put the value in the equation

$l\times v=y\times v_{M}$

$v_{M}=\dfrac{lv}{y}$

$v_{M}=\dfrac{v}{\dfrac{y}{l}}$

According to figure,

$\cos\theta=\dfrac{y}{l}$

Put the value of $\dfrac{y}{l}$

$v_{M}=\dfrac{v}{\cos\theta}$

Hence, The speed of mass will be $\dfrac{v}{\cos\theta}$.