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
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?
1 answer
1 answer