When a family of four with a total mass of 200 kg step into their1200-kg car, the car’s springs compress 3.0 cm.(a) What is the spring constant of the car’s springs, assumingthey act as a singlespring?(b) How far will the car lower if loaded with 300 kg rather than200 kg?
Question
When a family of four with a total mass of 200 kg step into their1200-kg car, the car’s springs compress 3.0 cm.(a) What is the spring constant of the car’s springs, assumingthey act as a singlespring?(b) How far will the car lower if loaded with 300 kg rather than200 kg?
Solution
Sure, let's solve this step by step.
(a) We can use Hooke's Law to find the spring constant. Hooke's Law states that the force exerted by a spring is equal to the displacement of the spring times the spring constant, or F = kx.
In this case, the force (F) is the weight of the family, which is their mass times gravity. The mass is 200 kg and gravity is approximately 9.8 m/s^2, so the force is 200 kg * 9.8 m/s^2 = 1960 N.
The displacement (x) is the amount the springs compress when the family gets in the car. This is given as 3.0 cm, but we need to convert this to meters because the standard unit of displacement in the SI system is meters. So, x = 3.0 cm = 0.03 m.
Now we can solve for the spring constant (k) using the formula F = kx:
1960 N = k * 0.03 m
Solving for k gives k = 1960 N / 0.03 m = 65,333.33 N/m.
(b) Now we want to know how far the car will lower if loaded with 300 kg rather than 200 kg. We can use the same formula, F = kx, but this time we know k and we're solving for x.
The force (F) is now the weight of the 300 kg load, or 300 kg * 9.8 m/s^2 = 2940 N.
Substituting the known values into the formula gives 2940 N = 65,333.33 N/m * x.
Solving for x gives x = 2940 N / 65,333.33 N/m = 0.045 m, or 4.5 cm. So the car will lower an additional 1.5 cm when loaded with 300 kg compared to 200 kg.
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