Jace asked in Science & MathematicsPhysics · 4 years ago

# A vertical spring (ignore its mass), whose spring stiffness constant is 900 N/m, is attached to a table and is compressed down 0.170 m.?

(a) What upward speed can it give to a 0.600 kg ball when released?

(b) How high above its original position (spring compressed) will the ball fly?

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• Jim
Lv 7
4 years ago

SPE = 1/2kx² = (0.5)(900)(0.170)² = 13.005 J

13.005 = KE = 1/2mV² = (0.5)(0.600)V² = 0.300V²

V² = 43.35

V = 6.58 m/s ANS (a)

time to reach max height = t = V/g = 6.58/9.81 = 0.67116 s

max height above top of uncompressed spring = h = 1/2gt² = (0.5)(9.81)(0.67116)² = 2.209 m

max height above top of compressed spring = h + 0.170 = 2.209 + 0.170 = 2.38 m ANS (b)

• Jace4 years agoReport

was pretty darn close, but maybe you switched some numbers? i got about .2 less for both answers from a friend who helped me. good work though.

• 4 years ago

Let’s use the following equation to determine the work that is done by the spring.

Work = ½ * k * d^2 = ½ * 900 * 0.170^2 = 13.005 N * m

As the spring expands 0.170 meter, the ball moves upward 0.170 meters. As this happens, its potential and kinetic energy increase.

∆ PE = 0.6 * 0.17 = 0.102 J

∆ KE = 13.005 – 0.102 = 12.903 J

½ * 0.6 * v^2 = 0.3 * v^2

0.3 * v^2 = 12.903

v^2 = 43.01

v = √43.01

This is approximately 6.56 m/s. As the ball rises to its maximum height, its velocity will decrease from 6.56 m/s to 0 m/s at the rate of 9.8 m/s each second. Use the following equation to determine this distance.

vf^2 = vi^2 + 2 * a * d

0 = 43.01 + 2 * -9.8 * d

19.6 * d = 43.01

d = 43.01 ÷ 19.6

This is approximately 2.19 meters. To determine its height above its original position, add 0.17 meter.