A 1300 kg car rolling on a horizontal surface has speed v = 65 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.9 m. What is the spring stiffness constant of the spring?
Solve this equation for k:
(1/2) k X^2 = (1/2) m V^2
(Initial kinetic energy) = (compressed spring potential energy)
V must be in m/s and X must be in meters.
k will be in N/m
To find the spring stiffness constant of the coiled spring, we can use the principle of conservation of mechanical energy.
The kinetic energy of the car before it strikes the spring is given by:
KE_initial = 0.5 * m * v^2
where m is the mass of the car and v is the initial velocity.
The potential energy of the compressed spring when the car comes to rest is given by:
PE_spring = 0.5 * k * x^2
where k is the spring stiffness constant and x is the distance compressed by the spring.
Since the car comes to rest after striking the spring, the initial kinetic energy is completely converted into the potential energy of the spring. Therefore, we can equate the two expressions:
0.5 * m * v^2 = 0.5 * k * x^2
Rearranging the equation:
k = (m * v^2) / x^2
Now, let's substitute the given values:
m = 1300 kg (mass of the car),
v = 65 km/h = (65 * 1000) / 3600 m/s (convert km/h to m/s),
x = 2.9 m (distance compressed by the spring).
Plugging these values into the formula:
k = (1300 * ((65 * 1000) / 3600)^2) / (2.9)^2
Now, let's calculate:
k ≈ 192416.67 N/m
Therefore, the spring stiffness constant of the coiled spring is approximately 192416.67 N/m.