Answer :
Final answer:
Using given data, various physics formulas, and Hooke's Law, the spring constant required for the car scenario provided is approximately 6789 N/m.
Explanation:
To calculate the spring constant k in the given scenario, we'll have to use a combination of formulas from kinetics and Hooke's law. Hooke's law states that the force exerted by a spring is equal to the displacement of the spring multiplied by the spring constant,
F = kx,
while the definition of acceleration is F = ma.
Through using the car's weight and the given maximum acceleration (4.0g, or about 39.2 m/s² as g is approximately 9.8 m/s²), one can find the force the spring should exert: 1200 kg * 39.2 m/s², which is approximately 47040 N.
Next, we need to find the maximum displacement of the spring (x), which is related to the initial speed of the car and the deceleration caused by the spring. Initial speed is given as 94 km/h, which translates to 26.11 m/s. Using the kinematic equation v² = u² + 2as, where u is the initial velocity, v is final velocity (which is zero in this case because the car comes to rest) and 'a' is the acceleration (negative in this case because it is deceleration), we solve for the displacement 's'.
s = -u² / (2a), this gives us the value of -6.93 m approximately.
Now having both F and x, we can go back to Hooke's Law and solve for k: k = F / x = 47040 N / 6.93 m which gives us k approximately equal to 6789 N/m. So, the spring constant required to bring a car of mass 1200 kg moving at 94 km/hr to complete rest with a maximum deceleration of 4.0g is approximately 6789 N/m.
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