A child\'s toy consists of m=33g monkey suspended from a spring of negligble mas

Question

A child's toy consists of m=33g monkey suspended from a spring of negligble 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= 14.5 cm as shown in the diagram. This toy is so adorable you pull the monkey down an additional d=5.7 cm from equilibrium and release it from rest, and smile with delight as it bounces playfully up and down.

so, m=33g ----> x=14.5cm -----> d=5.7cm

a) Using the given information, determine the spring constant, k, in Newtons per meter, of the spring.

b) Select free body diagram that represents the forces acting on the monkey as you are pulling it down, immediately before you let it go.

c) Calculate the potential energy, Ebottom, in joules stored in the strectched spring immediately before you release it

d) Assume that the system has zero gravitational potential energy at the lowest point of the motion. Derive an expression for the total mechanical energy, Eequilibrium, of the system as the monkey passes through the equilibrium position in terms of m, x, d, g, k, and the speed of the monkey, ve

e) Calculate the speed of the monkey, ve, in meters per second, as it passes through the equilibrium.

f) Derive an expression for the total machanical energy of the system as the monkey reaches the top of the motion, Etop, in terms of m, x, d, k, the maximum height above the bottom of the motion hmax, and the variables avaliable in the palette.

g) Calculate the maxiumum displacement, h, in centimeters above the equilibrium position that the monkey reaches.

Explanation / Answer

mass,m=33g=0.033 kg

streched distance,x=14.5cm=0.145m

Additional distance moved, d=5.7cm=0.057m

First we need to determine the force, Force,F=ma=0.033*0.057=0.00181 N

Force constant,k=-F/x=- 0.00181 N/0.145=-0.013 N/m

We know that the potential energy,PE=0.5kx2=0.5*0.013*0.1452=1.4*10-4 J

At equlibrium KE=PE

i.e 0.5mv2=0.5kx2

mv2=kx2

We get by substituting,

v=0.045 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.