1 Answer | Add Yours
Using Hooke's Law F = -kx, the spring constant of the spring can be determined.
When a person with a mass of 90 kg gets into the car, the springs sag by 12 cm. The force exerted by the mass of 90 kg is 90*9.8 = 882 N
882 = k*x = k*(0.12)
=> k = 882/0.12
=> k = 7350 N/m
The energy stored in the springs when the person with a mass of 90 kg enters the car is equal to (1/2)*k*x^2 = (1/2)*7350*(0.12)^2 = 52.92 J
It should be noticed that the potential energy in the spring is not the same as the loss in gravitational potential energy if a mass of 90 kg drops by 12 cm. This is due to the fact that the spring exerts a force in the opposite direction when the weight is applied.
We’ve answered 319,197 questions. We can answer yours, too.Ask a question