Question

A block of mass 20 kg is hung with the help of ideal
string, pulleys and spring (Spring constant
k = 1000 N/m) as shown in figure. If block is in
equilibrium position then extension in the spring will
be (g = 10 ms-2)
TO
20 kg
(1) 10 cm
B 5 cm
208 (2) 2.5 cm
0 (4) 15 cm​

Answer 1

Answer:

$x = 20 cm$

Explanation:

As we know that the block is hung by spring of ideal spring constant

$k = 1000 N/m$

here we know that at equilibrium we will have

$F_{net} = 0$

here we will have

$F_{net} = mg - kx$

so we have

$0 = mg - kx$

$0 = 20 (10) - 1000x$

$x = \frac{20\times 10}{1000}$

$x = 20 cm$

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Topic : Spring Force

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Answer 2

The extension in the spring will  be 20 cm.

Explanation:

It is given that,

Mass of the block, m = 20 kg

Spring constant of the spring, k = 1000 N/m

If block is in  equilibrium position then extension in the spring, then the weight of the spring is balanced by force due to spring such that,

$kx=mg$

$x=\dfrac{mg}{k}$

$x=\dfrac{20\times 10}{1000}$

x = 0.2 meters

or

x = 20 cm

So, the extension in the spring will  be 20 cm.

Learn more :

Spring constant

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