Asmall ball is attached to one end of a spring that has an unstrained length of 0.169 m. the spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.15 m/s. the spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0131 m. by how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

The ball moves around in circular motion. The centripetal force keeping it in circular motion is given by:

F = mv²/r

F = centripetal force, m = mass of ball, v = velocity of ball, r = radius of motion

The force the spring exerts on the ball is given by:

F = kΔx

F = spring force, k = spring constant, Δx = change of spring length

The spring provides the centripetal force that keeps the ball in circular motion, so set the spring force equal to the centripetal force:

kΔx = mv²/r

Let's do some algebra:

m/k = rΔx/v²

Now if we attach the same spring to the ceiling with the ball still fixed to one end and let the ball hang in static equilibrium, the spring force would balance the ball's weight:

kΔx' = mg

k = spring constant, Δx' = new change in spring length, m = mass, g = gravitational acceleration

Let's do some more algebra:

Δx' = gm/k

Substitute m/k with our previous result, rΔx/v²:

Δx' = grΔx/v²

Given values:

g = 9.81m/s²

r = 0.169m + 0.0131m = 0.1821m

Δx = 0.0131m

v = 3.15m/s

Plug in the values and solve for Δx':

Δx' = 9.81(0.1821)(0.0131)/3.15²

Δx' = 0.0024m

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2