Question

✓ Correct Correct answer is shown. Your answer 35933 N/m was either rounded differently or used a different number of significant figures than required for this part. Important: If you use this answer in later parts, use the full unrounded value in your calculations If you need to use the spring constant in subsequent parts, use the full precision value you calculated, only rounding as a final step before submitting your answer. Learning Goal: To understand the use of Hooke's law for a spring Hooke's law states that the restoring force on a spring when it has been stretched or compressed is proportional to the displacement 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 a means that F is equal to a constant times . 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 =-67 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 Part B After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average mass of 66 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 are the springs compressed? Enter the compression numerically in meters using two significant figures. View Available Hint(s) IVO AZO ? 2 = 6.8x10-1 Submit Previous Answers * Incorrect; Try Again; 8 attempts remaining The correct answer does not depend on: 210. Part C Complete previous part(s) Part D Complete previous part(s)

Answer 1

Let us inside here are total 28 in he taptap just people be for me addition of load. New added (74154 583425) ng (45+15+25) ug = 85kg Now 359 33 N/m. contanh spring Faux where a in the compression of sping due to me addition of me load. 7 ng 2 ur 2 21 mg ke 85 X 9.8 35933 ²0.023m But it consider the compression of he sping of the taptap with respect to that of without panenger and load then it ay will be (854 28X66) 35933 0.05379 m ~ 0.054 m CSScanned with CamScanner