Learning Goal: To understand the use of Hooke\'s law for a spring. Hooke\'s law

Question

Learning Goal: To understand the use of Hooke's law for a spring. Hooke's law states that lie restoring force F vector on a spring when it has been stretched or compressed is proportional to the displacement x bar 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 vector infinity x bar means that F vector is equal to a constant times x bar. 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 he written F vector k x bar. The minus sign indicates that the force is in the opposite direction to that or the springs 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. In 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 do or more passengers plus an assortment of chickens, goals, luggage, etc. Pulling this much into the back of a pickup truck 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 springs. Also for simplicity, assume that all tour 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 67 kg driver gets into an empty taptap to start the day's work. The springs compress 2.5 times 10^-2m. What s the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures. k = 2.6 times 10^4 N/m After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the drive, with an average mass of 67 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much ere the springs compressed? Enter the compression numerically in meters using two significant figures.

Explanation / Answer

Given K=2.6x104 N/m

Total weight = (27*67+3*15+3*5+25)*9.81=1894 kg *9.81 m/s2=18580.14N

Compression in spring= 18580.14(N)/(2.6x104 N/m)=0.714m

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