A child\'s toy consists of a m=31 g monkey suspended from a spring of negligible

Question

A child's toy consists of a m=31 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monkey is first hung on the spring and the system reaches equilibrium, the spring has stretched a distance of x= 16.3 cm, as shown in the diagram. This toy is so adorable you pull the monkey down an additional d=6.4 cm from equilibrium and release it from rest and smile with delight as it bounces playfully up and down.

a) Calculate the potential energy Ebottom in joules stored in the stretches spring immediately before you release it.

b) Calculate the speed of the monkey Ve in meters per second as it passes through equilibrium?

c) Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion Etop in terms of m, x, d, k, the maximum heigh above the bottom of the motion hmax and the variable avaiables in the palette?

d) Calculate the maximum displacement h in cm above the equilibrium position that the monkey reaches?

Explanation / Answer

when toy is at equilibrium position,

Fnet = k x - m g = 0

k(0.163) = (0.031)(9.8)

k = 1.864 N/m

(a) Spring PE = k x^2 / 2 = (1.864) (0.163 + 0.064)^2 /2

= 0.048 J

(b) at equilibrium position,

total Pe = (1.864)(0.163^2)/2 + (0.031)(9.8)(0.064)

= 0.044 J

Applying energy conservation,

0.048 + 0 = 0.044 + (0.031 v^2 / 2)

v = 0.51 m/s

(c) total mechanical energy = Gravittaional PE + spring PE + KE

= ( m g Hmax) + (k (Hmax - d - x)^2 / 2) + 0

= ( m g Hmax) + (k (Hmax - d - x)^2 / 2)

(d) Applying energy conservation,

( m g Hmax) + (k (Hmax - d - x)^2 / 2) = 0.048

(0.031)(9.8)(hmax) + (1.864)(hmax - 0.163 - 0.064)^2 / 2 = 0.048

0.3038 hmax + 0.932 hmax^2 - 0.423 hmax + 0.048 = 0.048

hmax ( 0.932 hmax - 0.1192) = 0

hmax = 0 Or 0.128 m

from equilibrium position = 12.8 - 6.4 = 6.4 cm .....Ans

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