A child\'s toy consists of a m = 27 g monkey suspended from a spring of negligib

Question

A child's toy consists of a m = 27 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 = 17.6 cm, as shown in the diagram. This toy is so adorable you pull the monkey down an additional d=7.5 cm from equilibrium and release it from rest, and smile with delight as it bounces playfully up and down. (a) Using the given information, determine the spring constant, k, in Newtons per meter, of the spring. (b) Select the free-body diagram that best represents the forces acting on the monkey as you are pulling it down, immediately before you let go. (c) Calculate the potential energy. E_bottom= in joules, stored in the stretched 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, E_equilibrium, 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, v_e. (e) Calculate the speed of the monkey, v_e, in meters per second, as it passes through equilibrium. (f) Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, E_top, in terms of m, x._ d._ k, the maximum height above the bottom of the motion, h_max. and the variables available in the palette. (g) Calculate the maximum displacement, h, in centimeters, above the equilibrium position, that the monkey reaches

Explanation / Answer

From the given question,

mass of monkey(m)=27 g=0.027kg

extension at equilibrium position(x)=17.6 cm=0.176m

at equilibrium position, mg=kx

0.027 *9.8=k *0.176

k=1.5

energy stored in lowest position(E1)= (1/2)(1.5)(17.6+7.5)2*10-4

Let at maximum displacement above equilibrium position be h

according to conservation of energy,

(1/2)k(h2)+ mgh=E1

(1/2)*(1.5)*h2+ 0.027*9.8*h=(1/2)(1.5)(17.6+7.5)2*10-4

solving we get

h=0.13m

=13 cm

Maximum displacement of monkey above equilibrium position is 13 cm.

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