Question

A block of mass 1 kg is connected with a spring of
spring constant 100 N/m which is connected to
the ceiling. If the block is released from the position
when the spring was at its natural length, what
will be the maximum elongation of the spring during
subsequent motion of block?

Answer 1

Explanation:

.2 meters is elongation of spring which is maximum and =( 2 mg)/ k

Answer 2

Answer:

A block of mass 1 kg is connected with a spring of  spring constant 100 N/m which is connected to  the ceiling. If the block is released from the position

when the spring was at its natural length, elongation of the spring is $0.09m$

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 $K=100N/m$

mass attached $m=1kg$

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

put the given values

⇒ $x=\frac{1*9.8}{100}=0.09m$