Question

A massless spring of spring constant 50 N/m in its unstretched position is hanged from the ceiling. A mass of 1 kg is attached and let go. Find the maximum elongation in the spring.

Answer 1

Answer:

A massless spring of spring constant 50 N/m in its upstretched position is hung from the ceiling. A mass of 1 kg is attached and let go, the elongation in the spring is $0.196m$

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 maximum elongation both the forces are balanced, that is $-Kx=mg$

From the question, we have

spring constant $K=50N/m$

mass attached $m=1kg$

maximum elongation of the spring $x=mg/K$

put the given values

⇒ $x=\frac{1*9.8}{50}=0.196m$