A child's toy consists of a m = 36 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 = 11.7 cm. 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.

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

Part (b) Draw the free-body diagram that best represents the forces acting on the monkey as you are pulling ti down, immediately before you let go.

Part (c) Calculate the potential energy, E
b
o
t
t
o
m
, in joules, stored in the stretched spring immediately before you release it.

Part (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
e
q
u
i
l
i
b
r
i
u
m
, 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
.

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

moya1316 moya1316 · Answer 2 · 2021-11-21T10:29:51+01:00

Newton's second law and Hooke's law allow us to find the results for the questions about the movement of the toy are:

a) The spring constant is: k = 3.0 N / m

b) free-body diagram in the adjoint

c) Potential energy is U = 4.54 10⁻² J

d) Mechanical energy at the equilibrium point is: Em = ½ m v² + m g x

Hooke's Law states that the force recovered in a spring is proportional to the displacement.

F = - k x

Where the bold letters indicate vectors, F is the force, k the spring conatant and x the displacement from the equilibrium point.

Newton's second law gives a relationship between the force, mass and the acceleration of bodies, in the special case that the acceleration is zero is called the equilibrium condition.

∑ F = 0

A free body diagram is a diagram of the forces without the details of the bodies, see attached.

a) They indicate the mass of the monkey is m = 0.036 kg, the displacement of the spring is x = 0.117 m. Let's write equilibrium condition.

F -W = 0

k x = m g

k = [tex]\frac{mg}{x}[/tex]

Let's calculate

k = [tex]\frac{0.036 \ 9.8 }{0.117}[/tex]

k = 3.0 N / m

b) In the attachment we have the requested free body diagram.

c) The potential energy of a spring is given by the relation:

U = ½ k x²

in the attachment we see the forces applied just before releasing the toy. Indicates additional stretch of x₀ = 5.7 cm = 0.057 m

[tex]x_{total}[/tex] = 11.7 + 5.7 = 17.4 cm

[tex]x_{total}[/tex] = 0.174 m

Let's calculate

U = ½ 3.0 0.174²

U = 4.54 10⁻² J

d) ask for mechanical energy when it passes through the equilibrium position.

As there is no friction in the system, energy is conserved at all points.

Let's look for the energy at the starting point, just before letting go of the toy.

Em = [tex]U_e[/tex]

Em = ½ k x²

The mechanical energy at the equilibrium point.

Em = K + [tex]U_g[/tex]

Em = ½ m v² + m g x

where x is the total spring displacement x = 0.174 m

Energy is conserved

½ k x² = ½ m v² + m g h

This expression can be used to find the speed of the monkey.

In conclusion, using Newton's second law and Hooke's law we can find the results for the questions about the movement of the toy are:

a) The spring constant is: k = 3.0 N / m

b) free-body diagram in the adjoint

c) Potential energy is U = 4.54 10-2 J

d) Mechanical energy at the equilibrium point is: Em = ½ m v² + m g x

Learn more here: brainly.com/question/18771646