Moderate -0.8 This is a straightforward application of Hooke's Law in equilibrium where tension equals weight. The calculation is direct: T = λx/l = mg gives 100x/0.5 = 20, so x = 0.1m. It's a single-step problem with standard recall of the elastic string formula, requiring no problem-solving insight, making it easier than average even for Further Maths mechanics.
1 In this question use \(g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
A particle of mass 2 kg is attached to one end of a light elastic string of natural length 0.5 metres and modulus of elasticity 100 N . The other end of the string is attached to the point \(O\).
Find the extension of the elastic string when the particle hangs in equilibrium vertically below \(O\).
Circle your answer.
0.01 m
0.1 m
0.2 m
0.4 m
1 In this question use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
A particle of mass 2 kg is attached to one end of a light elastic string of natural length 0.5 metres and modulus of elasticity 100 N . The other end of the string is attached to the point $O$.
Find the extension of the elastic string when the particle hangs in equilibrium vertically below $O$.
Circle your answer.\\
0.01 m\\
0.1 m\\
0.2 m\\
0.4 m
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2020 Q1 [1]}}