Easy -1.2 This is a direct substitution into the elastic potential energy formula EPE = λx²/(2l). Given EPE=4J, x=0.05m, l=0.5m, solving for λ requires only one algebraic step: λ = 2×EPE×l/x² = 1600N. It's a 1-mark multiple choice question testing formula recall with no problem-solving required, making it easier than average even for Further Maths.
A spring of natural length 50 cm and modulus of elasticity \(\lambda\) newtons has an elastic potential energy of 4 J when compressed by 5 cm.
Find the value of \(\lambda\)
Circle your answer.
[1 mark]
8 16 800 1600
A spring of natural length 50 cm and modulus of elasticity $\lambda$ newtons has an elastic potential energy of 4 J when compressed by 5 cm.
Find the value of $\lambda$
Circle your answer.
[1 mark]
8 16 800 1600
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2021 Q1 [1]}}