A 1000 kilogram car rolling on a horizontal surface has speed v= 45 kilometer per hour when it strikes a horizontal coil spring and is brought to rest in a distance of 2.2 meters. What is the spring stiffness constant of the spring?
To determine the spring stiffness constant of the spring, we can use the principle of conservation of mechanical energy. The mechanical energy of the car before striking the spring is converted into potential energy stored in the spring.
First, let's convert the speed of the car from kilometers per hour (km/h) to meters per second (m/s):
1 kilometer = 1000 meters
1 hour = 3600 seconds
So, the speed of the car in m/s is:
v = 45 km/h = (45 * 1000 m) / (3600 s) = 12.5 m/s
Next, we can use the equation for potential energy stored in a spring:
Potential Energy (PE) = (1/2) * k * x^2
Where:
PE is the potential energy stored in the spring
k is the spring stiffness constant (what we want to find)
x is the distance the spring is compressed (2.2 meters in this case)
Since the car comes to rest, its kinetic energy is completely converted into potential energy stored in the spring. Therefore, we can equate the initial kinetic energy of the car to the potential energy stored in the spring:
Initial Kinetic Energy = Final Potential Energy
(1/2) * m * v^2 = (1/2) * k * x^2
where:
m is the mass of the car (1000 kg)
v is the initial velocity of the car (12.5 m/s)
x is the distance the spring is compressed (2.2 meters)
Substituting the given values into the equation, we get:
(1/2) * 1000 kg * (12.5 m/s)^2 = (1/2) * k * (2.2 m)^2
62500 kg·m^2/s^2 = 1.21 * k
To find k, we can rearrange the equation:
k = (62500 kg·m^2/s^2) / (1.21 * (2.2 m)^2)
k ≈ 9770 N/m
Therefore, the spring stiffness constant of the spring is approximately 9770 N/m.