A 950 kg car rolling on a horizontal surface has speed v = 50 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?

initialKE = finalPE of spring

1/2 m v^2= 1/2 k x^2
solve for k

78

To find the spring stiffness constant, also known as the spring constant, we can use Hooke's Law and the concept of energy conservation.

1. Start by converting the speed from km/h to m/s:
Speed (v) = 50 km/h = (50 * 1000 m) / (3600 s) = 13.89 m/s

2. Calculate the initial kinetic energy of the car:
Kinetic Energy (KE_initial) = (1/2) * mass * (velocity)^2
= (1/2) * 950 kg * (13.89 m/s)^2

3. Since the car comes to rest, all the initial kinetic energy is transferred to the potential energy of the spring:
Potential Energy (PE_spring) = (1/2) * spring constant * (compression)^2

4. We are given the displacement (2.2 m) over which the car comes to rest, which is the compression of the spring.

5. Rearrange equation (3) to solve for the spring constant:
spring constant = (2 * PE_spring) / compression^2

But we need to find the potential energy of the spring first.

6. Substitute the expressions for kinetic energy and potential energy into equation (3):
(1/2) * 950 kg * (13.89 m/s)^2 = (1/2) * spring constant * (2.2 m)^2

7. Simplify the equation and solve for the spring constant:
spring constant = (950 kg * (13.89 m/s)^2) / (2 * (2.2 m)^2)

Calculate the right-hand side of the equation to find the spring constant.

Using these steps, you can find the spring constant for the given values.