Question

Natural length of a spring is 60 cm, and its spring
constant is 4000 N/m.Amass of 20 kg is hung from
it. The extension produced in the spring is, (take g =
9.8 m/s2):
(a) 4.9 cm
(b) 0.49 cm
(c) 9.4 cm
(d) 0.94 cm​

Answer 1

$\LARGE\underline{\underline{\sf Given:}}$

• Lenght of spring = 60cm

• Spring constant ( k) = 4000N/m

• Mass hung (m) = 20kg

$\LARGE\underline{\underline{\sf To\:Find:}}$

• Extension produced by spring (x) = ?

$\LARGE\underline{\underline{\sf Solution}}$

We know ,

$\LARGE{\boxed{\sf F=kx}}$

$\large\implies{\sf x=\dfrac{F}{k}}$

$\large\implies{\sf x=\dfrac{mg}{k} }$

$\large\implies{\sf x=\dfrac{2×9.8}{4000} }$

$\large\implies{\sf x=4.9\:cm }$

$\LARGE\underline{\underline{\sf Answer:}}$

Option (a) 4.9cm

•°• Extension produced in the spring is 4.9 cm

Answer 2

$\huge{ \boxed{ \boxed{ \mathfrak{solution}}}}$

$\LARGE{ \blue{\underline{\underline{\sf {Given:}}}}}$

1. Length of spring = 60cm
2. Spring constant ( k) = 4000N/m
3. Mass hung (m) = 20kg

$\LARGE{ \green{\underline{\underline{\sf { Required \: To\:Find:}}}}}$

Extension produced by spring (x) = ?

$\LARGE{ \pink{\underline{\underline{\sf {Answer : Option \: ' a '}}}}}$

We know ,

$\LARGE{\boxed{ \boxed{\sf F=kx}}}$

$\large\implies{\sf x=\dfrac{F}{k}}$

$\large\implies{\sf x=\dfrac{mg}{k} }$

$\large\implies{\sf x=\dfrac{2×9.8}{4000}}$

$\large\implies{\sf x=4.9\:cm}$

$\huge \orange{ \underline{ \underline{ \sf{Option \: (a) - 4.9}}}}$

Therefore , the extension offered by the spring is 4.9 cm .