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One of the greatest difficulties with setting up the baby bouncer is determining the right height above the floor so that the child can push off and bounce. Knowledge of physics can be really helpful here. If the spring constant k=5.0×102N/m, the baby has a mass m=11kg, and the baby's legs reach a distance d=0.15m from the bouncer, what should be the height h of the "empty" bouncer above the floor?

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.

Learn more about Spring force:

https://brainly.com/question/4291098

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