Final answer:
The spring stiffness constant required to stop a 1300-kg car moving at 65 km/h within 2.1 m is approximately 4300 N/m when calculated using the principle of conservation of energy.
Step-by-step explanation:
To find the spring stiffness constant of a spring that brings a car to rest, we can use the energy conservation principle. The car's kinetic energy at the moment of hitting the spring will be fully converted into the spring's elastic potential energy at the moment the car stops. Therefore, the equation representing this conservation is ½ mv2 = ½ kx2, where m is the mass of the car, v is the speed of the car, k is the spring constant, and x is the compression distance of the spring. Translating the car's speed to meters per second gives us 65 km/h = 18.0556 m/s. Using the car's mass (1300 kg) and the given compression distance (2.1 m), we can solve for k. After plugging in the values, we get k = (1300 kg × (18.0556 m/s)2) / (2 × (2.1 m)2) which gives a spring constant k to be about 4265.3 N/m. The spring stiffness constant of the spring is approximately 4300 N/m to two significant figures.