Question

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. The spring is then released shooting the spider up into the air. Of course the giant spider lands on Ms. Muffet while eating her candy and scares Ms. Muffet away. Okay, okay it actually ate her. (a) How much energy is stored in the compressed spring? 928 J (b) How much potential energy did the spider initially have while sitting on the spring? -62.72 J (c) What was the initial kinetic energy of the spider before the spring is released? 0 J (d) How high above the ground does the spider shoot before falling back to earth? 11.0367 m BONUS: What is the maximum speed of the spider during this process?

Answer 1

Answer:

a)  $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      $k_{e}$ = ½ k x²

      $k_{e}$ = ½ 2900 0.80²

      $k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     $E_{mf}$ = mg y

     Emo = $E_{mf}$

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     $E_{mf}$= ½ m v²

     Emo = $E_{mf}$

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s