## Mastering Physics Solutions: Baby Bounce with a Hooke

Baby Bounce with a Hooke

Part A = k = 2.5/0.005
Part B = 0.37 m

Solution Below:

One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or

F = -kx.

In its scalar form, this equation is simply

F = -kx.

The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation.

Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment.

Part A

The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer.

What is the spring constant of the spring being tested for the baby bouncer?

Remember that F = -kx. All you have to do is rearrange the equation:

F = -kx
-k = F/x

Drop the minus sign since “k” is always positive:

k = F/x

So you can take any pair of values from the table and find k, but since mastering physics wants a simplified answer in fractional form, use:

k = 2.5/0.005

k = 2.5/0.005

Part B

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 * 10^2 N, the baby has a mass m = 11 kg, and the baby’s legs reach a distance d = 0.15 m from the bouncer, what should be the height of the “empty” bouncer above the floor?

For this problem, just find the force of gravity on the baby and set it equal to the force on the spring. Start with gravity:

Fg = mg
Fg = 11 * 9.8
Fg = 107.8 N

Now plug this value into the formula for the force of the spring.

F = -kx
107.8 = -(500) * x
x = -0.2156

Subtract the length of the baby’s legs:

x = -0.2156 – 0.15
x = -0.3656

This is a negative value because the spring will be stretching *down* – so we know that the spring has to be 0.3656 m above the ground so that the baby’s feet will just touch the floor.

= 0.37 m

0.37 m

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