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The weight of the car causes compression in the springs. The length of each spring is reduced from its original length. Hooke’s Law can be used for compression as well as extension: F = kx where F = load, k = spring constant and x = compression. The weight of the car is 17 000 N. Each spring has a spring constant of 25...
See it on the paper →Use Fig. 3.2 to calculate the spring constant of the spring. spring constant = N / cm
See it on the paper →Which expression is equal to the spring constant?
See it on the paper →Spring A has a spring constant of 0.50 N / cm. Calculate the weight of the mass hanging from spring A. weight = N
See it on the paper →State Hooke’s law.
See it on the paper →Some candidates did not consider the original length of the spring, using the value of 12.0 cm for the extension. However, credit was available for the correct use of the formula.
— Examiner report, February/March 2024, 0654/42 Q3(a)(ii) (Cambridge)
A large majority of candidates were aware that it is the extension of the spring that must be considered, rather than the length. However, many selected option C, the reciprocal of the correct expression.
— Examiner report, May/June 2022, 0654/22 Q28 (Cambridge)
There was some confusion evident between the extension and the extended length, with some candidates using the extended length of 3.4 cm in their calculations rather than the extension of 1.2 cm. Candidates that did calculate the extension were generally able to calculate the weight correctly.
— Examiner report, February/March 2022, 0654/42 Q9(a)(i) (Cambridge)
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