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When a load of sand with mass 200 kg is placed in the back of a pickup truck the truck's rear lowers by 8.5 cm. Assuming the rear of the truck is supported by two identical springs, find the spring constant of those springs. How much work is done on the springs when the truck is loaded?

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Final answer:

The spring constant of the springs is 22752 N/m. The work done on the springs when the truck is loaded is 81.732 J.

Step-by-step explanation:

To find the spring constant of the springs, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the displacement of the truck's rear is 8.5 cm (or 0.085 m) when a load of sand with a mass of 200 kg is placed in the truck. To find the spring constant, we can rearrange the formula to solve for k:

k = F / x = (mg) / x = (200 kg)(9.8 m/s^2) / 0.085 m = 22752 N/m

The spring constant of the springs is therefore 22752 N/m.

The work done on the springs when the truck is loaded can be calculated using the formula for work: W = (1/2)kx^2. Since the displacement of the springs is 0.085 m, we can substitute the given values into the formula to calculate the work:

W = (1/2)(22752 N/m)(0.085 m)^2 = 81.732 J

Therefore, the work done on the springs when the truck is loaded is 81.732 Joules.

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