P5 Investigating Springs (page 206)

Question 1

How do you investigate the link between force and extension? (practical experiment)

Answer 1

Set up the apparatus as shown in diagram 1 (p.206)

Make sure you have plenty of extra masses, then measure the mass of each (with a mass balance) and calculate its weight (the force applied) using W = mg (p.202).

1) measure the natural length of the spring (when no load is applied) with a millimetre ruler clamped to the stand. Make sure you take the reading at eye level and add a marker (e.g. a thin strip of tape) to the bottom of the spring to make the reading more accurate.

2) Add a mass to the spring and allow it to come to rest. Record the mass and measure the new length of the spring. The extension is the change in length.

3) Repeat this process until you have enough measurements (no fewer than 6).

4) Plot a force-extension graph of your results (see graph on page 206). (it will only start to curve if you exceed the limit of proportionality, but don’t worry if yours doesn’t as long as you’ve got the straight line bit)

Question 2

You could do a quick piolt experiment first to check your masses are a good size, explain the process?

Answer 2

Using an identical spring to the one you’ll be testing, load it with masses one at a time up to a total of five. Measure the extension each time you add another mass.

Work out the increase in the extension of the spring for each of your masses. If any of them cause a bigger increase then the previous masses, you’ve gone past the spring’s limit of proportionality. If this happens, you’ll need to use smaller masses, or else you won’t get enough measurements for your graph.

Question 3

To check whether the deformation is elastic or inelastic, what do you do?

Answer 3

you can remove each mass temporarily and check the spring goes back to the previous extension.

Question 4

When the line of best fit is a straight line, what do this mean? (see graph on page 208).

Answer 4

it means there is a linear relationship between force and extension (they’re directly proportional, see page 205).

F = ke, so the gradient of the straight line is equal to k, the spring constant.

Question 5

When the line begins to bend on the graph on page 206, what is the relationship - linear or non-linear, and why?

Answer 5

it is now non-linear between force and extension - the spring stretches more for each unit increase in force.

Question 6

As long as a spring is not stretched past its limit of proportionality, the work done is stretching (or compressing) a spring can be found using what equation?

Answer 6

Ee = ½ke²

Ee - Elastic potential energy (J)

k - Spring constant (N/m)

e - Extension (m)

see equation on page 206.

The energy in the elastic potential energy store of a stretched spring is equal to the area under a force-extension graph up to that point. (see diagram on page 206).

Question 7

For elastic deformation, this formula can be used to calculate what?

Answer 7

energy stored in a springs elastic potential energy store. It’s also the energy transferred to the spring as it’s deformed (or transferred by the spring as it returns to its original shape).

Question 8

Remember you can only use the gradient to find the spring constant if the graph is linear or non-linear?

Answer 8

when the graph is linear (a straight line).

Question 9

A spring with a spring constant of 40 N/m extends elastically by 2.5cm Calculate the amount of energy stored in its elastic potential energy store? (3 marks)

Answer 9

2.5cm = 0.025 m (1 mark)
Ee = ½ke² = ½ x 40 (0.025)² (1 mark)
= 0.0125 J (1 mark)