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Forces and Newton’s Laws: Question 8

Syllabus 1.5.1

Structured Extended 8 marks

A technician tests two different springs, P and Q, by hanging various loads from each and measuring the extension produced. Both springs obey Hooke's law over the full range of loads shown. The results are given in the table below.

Load, FF / N 0 2.0 4.0 6.0 8.0
Extension of spring P, xPx_P / cm 0 0.8 1.6 2.4 3.2
Extension of spring Q, xQx_Q / cm 0 1.5 3.0 4.5 6.0

For each spring, a graph of load, FF (vertical axis), against extension, xx (horizontal axis), gives a straight line through the origin, and the gradient of this line is equal to the spring constant.

(a) Using two points from the table for spring P, calculate the gradient of its load-extension line, and hence state its spring constant kPk_P, in N/cm. [2]

(b) Using two points from the table for spring Q, calculate the gradient of its load-extension line, and hence state its spring constant kQk_Q, in N/cm. [2]

(c) State which spring, P or Q, is stiffer, and explain how the spring constants found in (a) and (b) show this. [2]

(d) Assuming spring P continues to obey Hooke's law beyond a load of 8.0 N8.0\text{ N}, calculate the extension it would produce for a load of 11.0 N11.0\text{ N}. [2]

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Worked solution

Part (a): Spring constant of spring P

Since the load-extension line passes through the origin and is straight, the gradient is the same anywhere along it. Using the points (2.0 N,0.8 cm)(2.0\text{ N}, 0.8\text{ cm}) and (8.0 N,3.2 cm)(8.0\text{ N}, 3.2\text{ cm}) for spring P:

kP=ΔFΔx=8.0 N−2.0 N3.2 cm−0.8 cm=6.0 N2.4 cmk_P = \frac{\Delta F}{\Delta x} = \frac{8.0\text{ N} - 2.0\text{ N}}{3.2\text{ cm} - 0.8\text{ cm}} = \frac{6.0\text{ N}}{2.4\text{ cm}}

kP=2.5 N/cmk_P = \boxed{2.5\text{ N/cm}}

Part (b): Spring constant of spring Q

Using the points (2.0 N,1.5 cm)(2.0\text{ N}, 1.5\text{ cm}) and (8.0 N,6.0 cm)(8.0\text{ N}, 6.0\text{ cm}) for spring Q:

kQ=ΔFΔx=8.0 N−2.0 N6.0 cm−1.5 cm=6.0 N4.5 cmk_Q = \frac{\Delta F}{\Delta x} = \frac{8.0\text{ N} - 2.0\text{ N}}{6.0\text{ cm} - 1.5\text{ cm}} = \frac{6.0\text{ N}}{4.5\text{ cm}}

kQ=1.3 N/cm (2 s.f.)k_Q = \boxed{1.3\text{ N/cm}} \text{ (2 s.f.)}

Part (c): Comparing the stiffness of the springs

The spring constant tells you how much force is needed to produce each centimetre of extension. A larger spring constant means a stiffer spring, since a bigger force is needed for the same extension.

Since kP=2.5 N/cmk_P = 2.5\text{ N/cm} is larger than kQ≈1.3 N/cmk_Q \approx 1.3\text{ N/cm}, spring P is stiffer than spring Q: for the same load, spring P stretches less than spring Q (as shown directly in the table).

Part (d): Extrapolating spring P’s extension

Rearranging k=F/xk = F/x to make extension the subject, and using spring P’s spring constant found in part (a):

kP=Fx⇒x=FkPk_P = \frac{F}{x} \quad\Rightarrow\quad x = \frac{F}{k_P}

x=11.0 N2.5 N/cmx = \frac{11.0\text{ N}}{2.5\text{ N/cm}}

x=4.4 cmx = \boxed{4.4\text{ cm}}

Final answers

  • (a) kP=k_P = 2.5 N/cm2.5\text{ N/cm}
  • (b) kQ≈k_Q \approx 1.3 N/cm1.3\text{ N/cm}
  • (c) Spring P is stiffer, since it has the larger spring constant
  • (d) Extension of spring P at 11.0 N11.0\text{ N} == 4.4 cm4.4\text{ cm}