# A child's toy consists of a m=31 g monkey suspended from a spring of negligible mass and spring...

## 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 height 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?

## Energy Conservation:

The mechanic energy is conserved if there is no non-conservative force, such as friction or air drag, in a system.

{eq}\begin{align} E_i & = E_f\\ KE_i + PE_i & = KE_f + PE_f \end{align} {/eq}

The kinetic energy of an object is

{eq}KE = \dfrac 12 m v^2 {/eq}

The potential energy of an object is

{eq}PE = mgh {/eq}

The potential energy of the spring is

{eq}PE = \dfrac 12 k x^2 {/eq}

Part a).When the toy is in equilibrium, it is stretched x = 16.3 cm. The spring force is equal to the weight. We have

{eq}\begin{align} kx & = mg\\ k & = \dfrac{mg}{x}\\ k & = \dfrac{0.031*9.8}{0.163}\\ k & = 1.8638 \rm \ N/m \end{align} {/eq}

As the spring is streched another d = 6.4 cm, there is no kinetic energy. Set the height H as zero at the equilibrium position. The initial energy before you release it is

{eq}\begin{align} E_i & = KE_i + PE_i\\ & = \dfrac 12 m v_i^2 + PE_i\\ & = 0 + PE_i\\ & = Ebottom\\ & = \dfrac 12 k (x + d)^2 + mgH_i\\ & = \dfrac 12 k (x + d)^2 + mg(-d)\\ & = \dfrac 12 k (x^2 + 2 xd + d^2) - mgd\\ & = \dfrac 12 k x^2 + kxd + \dfrac 12 k d^2 - mgd\\ & = \dfrac 12 k x^2 + \dfrac 12 k d^2\\ & = \dfrac 12 k (x^2 + d^2)\\ & = \dfrac 12 1.8638* (0.163^2 + 0.064^2)\\ & = \boxed{\bf 0.0286 \ \rm J} \end{align} {/eq}

Part b). The energy is conserved and we can find the kinetic energy as it passes through equilibrium and solve for the speed Ve.

{eq}\begin{align} E_i & = E_e \\ Ebottom & = E_e\\ \dfrac 12 k x^2 + \dfrac 12 k d^2 & = \dfrac 12 k x^2 + \dfrac 12 m Ve^2\\ \dfrac 12 k d^2 & = \dfrac 12 m Ve^2\\ k d^2 & = m Ve^2\\ Ve^2 & = \dfrac{k d^2}{m}\\ Ve & = \sqrt{\dfrac{k d^2}{m}}\\ Ve & = \sqrt{\dfrac{1.8638 * 0.064^2}{0.031}}\\ Ve & = \boxed{\bf 0.246 \ \rm m/s} \end{align} {/eq}

Part c). At the top of the motion, the monkey would stop and the velocity would be zero. The maximum height is measured from the bottom of the motion and h is measured from the equilibrium. We have

{eq}hmax = h + d {/eq}

The kinetic energy is zero and the total energy can be found as :

{eq}\begin{align} Etop & = KE_t + PE_t\\ & = 0 + PE_t\\ & = \dfrac 12 k (x - H_t )^2 + mgH_t\\ & = \dfrac 12 k (x - h)^2 + mgh\\ & = \dfrac 12 k (x^2 - 2xh + h^2) + kxh\\ & = \dfrac 12 k (x^2 + h^2)\\ & = \boxed{\bf \dfrac 12 k (x^2 + (hmax - d)^2)} \end{align} {/eq}

Part d). The total mechanic energy is conserved. We can solve for the maximum displacement h:

{eq}\begin{align} E_i & = Etop\\ \dfrac 12 k (x^2 + d^2) & = \dfrac 12 k (x^2 + h^2)\\ x^2 + d^2 & = x^2 + h^2\\ d^2 & = h^2\\ h & = d\\ h & = \boxed{\bf 6.4 \ \rm cm} \end{align} {/eq}