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## A 1.01 kg mass is attached to a spring of force constant 10.1 N/cm and placed on a frictionless surface. By how much will the spring stretch

Question

A 1.01 kg mass is attached to a spring of force constant 10.1 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.515 m at a rate of 1.84 revolutions per second

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Physics
7 months
2021-07-27T11:43:02+00:00
2021-07-27T11:43:02+00:00 1 Answers
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## Answers ( )

Answer:6.88 cmExplanation:Given that:

mass

mof the object = 1.0 kgForce constant of the spring (k) = 10.1 N/cm = 10.1 × (100) N/m

= 1010 N/m

Radius of the circular path R = 0.515 m

Since 1 revolution = 2 π radian

angular speed of the object

ω= 1.84 rev/s = 1.84 × ( 2 π) rad/s= 11.56 rad/s

Amount of the spring that stretches = X???

However, the amount of the force the spring exerts on the object due to the stretching is directly equal to the centripetal force required by the object for the circular motion:

SO;

X = 0.0688 m

Hence, the amount of the spring stretches =

6.88 cm