The Physics
Hypertextbook
Opus in profectus

# Springs

## Practice

### practice problem 1

A chest expander is a type of exercise device made from three or four large parallel springs connected to a pair of handles. The springs offer muscle resistance in a way that is similar to lifting weights. One such device broke and was donated to a physics teacher. The teacher attached one of the springs to a ceiling beam in a classroom at started loading it with 1 kg masses. A pair of vertically stacked meter sticks were placed parallel to the loaded spring and the position of the bottom of the spring was recorded. Use this data to determine the spring constant of this spring.

#### solution

Hooke's law states that the force (F) on a spring is proportional to its extension (x).

F = kx

The proportionality constant in Hooke's law is called the spring constant (k).

 k = F ∆x

The data given in this experiment were mass and position, not force and extension. The force in this experiment comes from the weight of the suspended masses. We have an equation for weight.

W = mg

Extension means change in length from some relaxed, natural, or unstressed length. Subtracting out the original position will give us the extension.

x = x − x0

Compute these two quantities and change the extension from centimeters to meters (since that's the preferred SI unit). Compile the results in a table like the one below.

Chest expander
mass
(kg)
position
(cm)
force
(N)
extension
(m)
0 026.5 00.0 0.000
1 040.0 09.8 0.135
2 054.0 19.6 0.275
3 068.5 29.4 0.420
5 097.5 49.0 0.710
6 112.5 58.8 0.860
7 127.5 68.6 1.010
8 143.5 78.4 1.170

Now graph it. Put force on the y-axis and extension on the x-axis. Since slope is…

 m = ∆y ∆x

the slope of a line of best fit will give us the spring constant. This spring from a chest expander has a spring constant of…

k = 67 N/m

### practice problem 2

A teacher presses down on a spring loaded pop-up toy until the suction cup part of it catches. The teacher's class waits in anticipation. Without warning the suction breaks and the toy does what it's supposed to do — it pops up. The teacher records the toy's mass, its height height before and after being compressed, and the maximum height it attains after popping up.
Pop-up toy
mass
of toy
uncompressed
height
compressed
height
maximum
height
16 g
0.016 kg
11cm
0.11 m
7 cm
0.07 m
92 cm
0.92 m
Assume that the trajectory of the toy was nearly vertical and that air resistance and other forms of friction are negligible and determine the following quantities…
1. the gravitational potential energy of the toy at its highest point relative to the table top
2. the kinetic energy of the toy as it left the table
3. the speed of the toy as it left the table
4. the elastic potential energy of the toy after it was compressed
5. the spring constant of the spring inside the toy

#### solution

1. Gravitational potential energy is found with the basic equation.

 ∆Ug = mg∆h∆Ug = (0.016 kg)(9.8 m/s2)(0.92 m)∆Ug = 0.144 J
2. The kinetic energy of the toy as it left the table is the same as the potential energy it had at the top of its flight. Energy changed form, but was conserved. There is nothing to calculate here.

K = 0.144 J

3. Use the kinetic energy to determine the speed of the toy as it left the table.

 v = √ 2K ⇐ K = ½mv2 m

Be sure to use SI units.

 v = √ 2K m
 v = √ 2(0.144 J) (0.16 kg)
 v = 4.24 m/s
4. The elastic potential energy of the compressed toy is the same as the kinetic energy it had when it took off. Once again, energy was conserved and there's nothing to calculate.

Us = 0.144 J

5. Use the elastic potential energy equation to find the spring constant.

 k = 2Us ⇐ Us = ½kx2 x2

Watch out for non SI units. Don't let them interfere with a proper solution.

 k = 2Us x2
 k = √ 2(0.144 J) (0.04 m)2
 k = 180 N/m

### practice problem 3

Write something different.