Question

A small ball is attached to one end of a spring that has an unstrained length of 0.175 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.86 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0140 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Answer 1

Answer:

x(spring stretch)=1.74×10⁻³ meter

Explanation:

Given Data

l (length)=0.175 m

v (speed)= 3.86 m/s

x₁(stretch)=0.0140 m

x(spring stretch)=?

Solution

Centripetal force=Spring Constant

$\frac{mv^{2} }{l}=kx_{1}\\ k=(1/x_{1} )\frac{mv^{2} }{l}\\ k=(1/0.014)(\frac{m(3.86)^{2} }{(0.175+0.014)} )\\k=m*5630.99$

as

Spring force = gravitational force

$kx=mg\\x=\frac{mg}{k}\\ x=\frac{m(9.8)}{m*5630.99}\\ x=\frac{9.8}{5630.99}\\ x=1.74*10^{-3} meter$