Asked by wanda
What should be the spring constant k (in N/m) of a spring designed to bring a 1105-kg car to rest from a speed of 69-km/h so that the occupants undergo a maximum acceleration of 3.8g's?
Answers
Answered by
Damon
F = k x
potential energy in spring = .5 k x^2
kinetic energy in car = .5 m v^2
so in the end
k Xmax^2 = m Vi^2 = 1105 (69000/3600)^2
Fmax = m Amax
k Xmax = 1105 (3.8*9.81)
so
Xmax = 1105(3.8*9.81)/k = 41192/k
-------------------
k(1.7*10^9)/k^2 = 406,000 =4.06*10^5
4.06 k = 1.7*10^4
k = 4187 N/m
check my arithmetic
potential energy in spring = .5 k x^2
kinetic energy in car = .5 m v^2
so in the end
k Xmax^2 = m Vi^2 = 1105 (69000/3600)^2
Fmax = m Amax
k Xmax = 1105 (3.8*9.81)
so
Xmax = 1105(3.8*9.81)/k = 41192/k
-------------------
k(1.7*10^9)/k^2 = 406,000 =4.06*10^5
4.06 k = 1.7*10^4
k = 4187 N/m
check my arithmetic
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.