Question

If a mass of 10 kg attached to a vertical spring stretches the spring 2 m from its original equilibrium position, what is the spring constant?

Answer 1

Answer:

49 N/m

Explanation:

Here we use, the simple Spring formula and g is 9.8ms-² which is almost equal in every region of earth.

Hope u get it...

Answer 2

Answer:

If a mass of 10 kg attached to a vertical spring stretches the spring 2 m from its original equilibrium position, then the spring constant is $49N/m$

Explanation:

• The restoring force on a spring of spring constant (K) when elongated to a distance (x) is given as $F=-Kx$

• Due to the mass (m) attached to the spring the gravitational force on the mass is  $F=mg$

        where $g=9.8m/s^{2}$(acceleration due to gravity)

• at equilibrium condition  both the forces are equal, that is $-Kx=mg$

From the question, we have

mass attached $m=10kg$

stretched length $x=2m$

spring constant $K=mg/x$

put the given values

⇒$K=\frac{10*9.8}{2} =49N/m$