Weight G=m×g=0.55×9.8N
G=k×dx, where dx=2cm is the stretch.
dx=2cm=0.02m
m×g=k×dx
k=m×g/dx=0.55×9.8/0.02=269.5N/m
Answer: k=269.5N/m
The value of spring constant If a mass of 0.55kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium is k=269.5N/m
The spring constant is defined as it is the ratio of force applied on the spring and the deflection of the spring.
The weight of the mass will be
F=m×g=0.55×9.8N
Spring force will be given as
F=k×dx, w
here dx=2cm is the stretch.
dx=2cm=0.02m
Since the weight of the mass will be equal to the force applied on the spring.
m×g=k×dx
k=m×g/dx=0.55×9.8/0.02=269.5N/m
k=269.5N/m
Hence value of spring constant If a mass of 0.55kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium is k=269.5N/m
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