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The springs of a 1200 kg car sag 12 cm when a 90 kg person gets in the car. Determine...

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qasenior | (Level 1) Salutatorian

Posted June 23, 2012 at 3:13 AM via web

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The springs of a 1200 kg car sag 12 cm when a 90 kg person gets in the car. Determine the spring constant and the total energy stored in the spring. 

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 24, 2012 at 4:27 PM (Answer #1)

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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.

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