A 900 kg car rolling on a horizontal surface has speed v = 55 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
the KE of the car= PE of the spring
1/2 m v^2=1/2 k x^2
solve for k
velocity has to be in m/s
To find the spring stiffness constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The formula for the force exerted by a spring is given by:
F = -kx
Where:
- F is the force exerted by the spring
- k is the spring stiffness constant
- x is the displacement of the spring from its equilibrium position
In this case, the car is brought to rest in a distance of 2.2 m. Since the car is initially moving and brought to rest, the displacement of the spring will be equal to the distance it takes for the car to stop.
So, x = 2.2 m
Now, we need to calculate the force exerted by the spring.
The initial kinetic energy of the car is given by:
KE = (1/2)mv^2
Where:
- KE is the kinetic energy
- m is the mass of the car
- v is the velocity of the car
Given that the mass of the car is 900 kg and the velocity is 55 km/h, we need to convert the velocity to m/s.
1 km/h = 0.2778 m/s
So, v = 55 km/h * 0.2778 m/s = 15.2778 m/s
Plugging the values into the equation:
KE = (1/2) * 900 kg * (15.2778 m/s)^2 = 207039.36 J
The work done by the spring is equal to the change in kinetic energy:
Work = KE = (1/2)kx^2
Since the car is brought to rest, the final kinetic energy is zero, so the work done is equal to the initial kinetic energy:
(1/2)kx^2 = 207039.36 J
Now, we can solve for the spring stiffness constant k:
k = (2 * Work) / x^2
k = (2 * 207039.36 J) / (2.2 m)^2
k = 168000 N/m
Therefore, the spring stiffness constant of the spring is 168000 N/m.