5.7 Required practical: force and extension
8 exam questions worth 19 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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State the independent variable and the dependent variable in this investigation.
Mark scheme
independent: the force / weight / mass added (1); dependent: the extension of the spring (1)
Model answer
The independent variable is the force (the weight of the masses). The dependent variable is the extension of the spring.
A mass of 0.20 kg hangs from a spring. Calculate the force on the spring. (g = 9.8 N/kg)
Mark scheme
W = m g = 0.20 × 9.8 (1); 1.96 N (1)
Model answer
W = m g = 0.20 × 9.8 = 1.96 N.
A spring has a natural length of 15.0 cm. With a load, its length is 21.5 cm. Calculate the extension.
Mark scheme
6.5 cm
Model answer
Extension = 21.5 − 15.0 = 6.5 cm.
Describe how the student should reduce parallax error when reading the length of the spring.
Mark scheme
read the ruler at eye level / use a set square so the line of sight is at right angles to the ruler (1); keep the pointer close to the ruler (1)
Model answer
Read the ruler with the eye level with the pointer, or use a set square. Keep the pointer close to the ruler.
The student removes the masses one at a time and records the lengths again. Explain why.
Mark scheme
to check that the spring returns to its original length (1); which shows whether the deformation was elastic / the limit of proportionality was not exceeded (1)
Model answer
To check that the spring returns to the same lengths, which shows the stretching was elastic and the limit of proportionality was not passed.
The graph of force against extension is a straight line through the origin. It passes through the point (0.080 m, 4.0 N). Calculate the spring constant.
Mark scheme
k = F ÷ e (1); 4.0 ÷ 0.080 (1); 50 N/m (1)
Model answer
k = F / e = 4.0 / 0.080 = 50 N/m.
The line on the graph is a straight line through the origin for small forces but curves for large forces. Explain what this shows.
Mark scheme
force is proportional to extension for small forces (1); for large forces the spring has gone past its limit of proportionality (1)
Model answer
For small forces the extension is directly proportional to the force. At large forces the spring has gone past the limit of proportionality.
Describe a method to investigate how the extension of a spring depends on the force applied. Include how you would make sure the results are accurate.
Mark scheme
measure the natural length with a ruler (1); add masses to the spring one at a time and measure the new length each time (1); calculate force from the weight of masses / extension from the lengths (1); use a set square / read at eye level, wait for spring to stop moving (1); repeat and find a mean / plot force against extension (1)
Model answer
Hang a spring from a clamp stand with a vertical ruler beside it. Measure the natural length. Add masses one at a time, waiting for the spring to stop moving, and record the length each time, reading the ruler at eye level. Calculate the extension and the force, repeat, and plot a graph of force against extension.
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