To solve this problem we will apply the energy conservation theorems for which the elastic potential energy is equivalent to the kinetic energy, this is
[tex]KE = PE[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2} kx^2[/tex]
Here,
m = Mass
v = Velocity
k = Spring constant
x = Displacement, our values are given as,
[tex]KE = 40,000 J,[/tex]
[tex]v = 7 m/s[/tex]
[tex]x = 0.3 m.[/tex]
Replacing we have that
[tex]40000 = \frac{1}{2} k (0.3)^2[/tex]
[tex]k = \frac{2*40000}{0.3^2}[/tex]
[tex]k = 888889N/m[/tex]
Therefore the bumper's spring constant is 888889N/m
Answer:
k = 888,888.89N/m
k = 8.9 × 10^5 N/m
Therefore, the bumper's spring constant is 8.9 × 10^5 N/m
Explanation:
Given that energy wasn't lost from the point when the car had 40,000 J of KE to when it impacted the wall, i.e. the energy was perfectly conserved (assumption)
The kinetic energy of the car would be converted to the Elastic potential energy of the bumper spring. According to the law of conservation of energy.
K.E = Elastic P.E = (1/2)kx^2. .....1
Where;
k = bumper spring constant.
x = compression length of spring.
From equation 1
k = 2K.E/x^2
K.E = 40,000J
x = 0.30m
k = 2(40000)/(0.30^2)
k = 80000/0.09
k = 888,888.89N/m
k = 8.9 × 10^5 N/m
Therefore, the bumper's spring constant is 8.9 × 10^5 N/m
