Answer: The change in elastic potential energy is 3,045 J.
Explanation: The change in elastic potential energy of the suspension between the maximum force before the spring breaks and when no additional force is being exerted on the car beside the Earth can be calculated using the formula:
ΔPE = 1/2k(x2^2 - x1^2)
where ΔPE is the change in elastic potential energy, k is the spring constant, x1 is the initial displacement of the suspension from its equilibrium position, and x2 is the maximum displacement of the suspension from its equilibrium position.
Assuming that the spring follows Hooke’s law, which states that the force exerted by a spring is proportional to its displacement from its equilibrium position, we can calculate the spring constant using:
k = F/x
where F is the maximum force exerted on the car in the vertical direction and x is the maximum displacement of the suspension from its equilibrium position.
Using these formulas and given that the mass of the car is roughly 2000kg and that the car suspension has an initial displacement of 8.5 cm from its equilibrium position, we can calculate that:
x2 = 8.5 cm + 6 cm = 14.5 cm
x1 = 8.5 cm
x = 14.5 cm - 8.5 cm = 6 cm
F = 1.4x10^4 N
k = F/x = (1.4x10^4 N) / (6 cm) = 2.33x10^5 N/m
ΔPE = 1/2k(x2^2 - x1^2) = (0.5)(2.33x10^5 N/m)((0.145 m)^2 - (0.085 m)^2) = 3,045 J
Therefore, the change in elastic potential energy of the suspension between the maximum force before the spring breaks and when no additional force is being exerted on the car beside Earth is 3,045 J.
Hope this helps, and have a great day!