Question

Hooke's law states force f in spring extended by a length is given by f=-kx according to newton's 2nd law f=ma calculate the dimensions of spring constant k

Answer 1

The dimensions of spring constant k is [MT⁻²]

Explanation:

Force F in spring extended by a length x is given by

$F=-kx$

According to the Newton's second law

$F=ma$

Where m is mass and a is acceleration

Thus we can write

$kx=ma$

$\implies k=\frac{ma}{x}$

$\implies \text{Dimensions of k}=\frac{\text{Dimension of m}\times\text{Dimensions of a}}{\text{Dimension of x}}$

$\implies \text{Dimensions of k}=\frac{[M][LT^{-2}]}{[L]}$

$\implies \text{Dimensions of k}=[MT^{-2}]$

Hope this answer is helpful.

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