Question

When a load of 10 kg is suspended on a metallic
wire, its length increase by 2 mm. The force
constant of the wire is
(1) 3 x 104 N/m (2) 2.5 x 103 N/m
(3) 5 x 104 N/m (4) 7.5 x 103 N/m

Answer 1

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

• Mass of load (m) = 10kg

• Lenght increase by (x) = $\sf{2×10^{-3} \:m}$

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

• Force constant ( k )

$\Large\underline{\underline{\sf Formula\: Used:}}$

$\large{\boxed{\boxed{\sf \pink{F=kx }}}}$

$\Large\underline{\underline{\sf Solution:}}$

$\Large{\sf F=kx }$

$\implies{\sf mg=kx}$

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

$\implies{\sf k=\dfrac{10×10}{2×10^{-3}}}$

$\implies{\sf \red{k=5×10^4\:N/m}}$

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

Option (3) 5 × 10⁴ N/m

⛬ Force Constant (k) is $\bf{5×10^4\:N/m}$

Answer 2

The force constant of the wire is $5\times 10^4\ N/m$.

Explanation:

Given that,

Mass of the load, m = 10 kg

Stretching in the spring due to weight, x = 2 mm = 0.002 m

We need to find the force constant of the wire. As the load is suspended on a wire, the force of gravity is balanced by the force acting on the spring such that,

$mg=kx$

k is the force constant of the wire

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

$k=\dfrac{10\times 10}{0.002}$

$k=5\times 10^4\ N/m$

So, the force constant of the wire is $5\times 10^4\ N/m$. Hence, this is the required solution.

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