To understand the sue of Hooke\'s law for a spring. Hooke\'s law states that the

Question

To understand the sue of Hooke's law for a spring. Hooke's law states that the restoring force F on a spring when it has been stretched or compressed is proportional to the displacement x of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed. Recall that F infinity x means that F is equal to a constant times x. For a spring, the proportionality constant is called the spring constant and denoted by k. The spring constant is a property of the spring and must be measured experimentally. The larger the value of k, the stiffer the spring. In equation form, Hooke's law can be written F = -kr. The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by F = kx, where x is the magnitude of the displacement. The Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goes, luggage, etc. Putting this much into the back of the pickup puts quite a large load on the truck springs. A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the spring. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck. A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.4 Times 10^-2 m. What is the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in netwons per meter using two significant figures.

Explanation / Answer

(A) F = mg = (70 kg) (9.8 m/s2)

F = 686 N

According to Hooke's law

F = kx

k = F/x

k = (686 N) / (2.4 x 10-2 m)

k = 28583.33 N/m

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