Respuesta :
Answer:36.56 cm
Explanation:
Given
Spring constant of spring [tex](k)=5\times 10^2 N/m[/tex]
mass of baby=11 kg
d= baby leg from bouncer to ground
At Equilibrium
mg=kx
where x= compression
[tex]x=\frac{11\times 9.8}{500}=21.56 cm[/tex]
So we need to set bouncer height =21.56+15=36.56 cm such that during equilibrium baby legs are just on ground
The height, h of the "empty" bouncer above the floor should be 0.366m.
What is the spring force?
Spring force is the product of the spring constant and the displacement of the spring.
We know that when the baby will jump on the bouncer the force that will be applied back by the bouncer is equal to the force applied by the baby on the bouncer. therefore,
Force applied by the bouncer = force applied by the baby
We know that the force applied by the baby is the weight of the baby, thus,
Force applied by the bouncer = weight of the baby
[tex]kx = mg\\\\5.0 \times 10^2 \times x = 11 \times 9.81 \\\\x = \dfrac{11 \times 9.81}{5 \times 10^2}\\\\x = 0.216\rm\ m[/tex]
As the displacement of the bouncer is 0.216 m, also, we know that the baby's legs reach a distance d=0.15m from the bouncer. thus, the height h of the "empty" bouncer above the floor should be 0.216 m + 0.15 m.
The height h of the "empty" bouncer = 0.216m + 0.15m
= 0.366 m
Hence, the height h of the "empty" bouncer above the floor should be 0.366 m.
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