Correct Answer to Question 9

Since the board reaches the floor, the length of the spring is X = 2.44 - 1.98 m. The unstrained length is 0.305 m, so the spring stretches an additional distance X = X - 0.305 m, when the board is hanged from the string.

Thus the board exerts a force of magnitude F = k X on the spring. Since the weight of the board is exerting the force on the spring, we can also write:

k X = m g

where m is the mass of the board (10.4 kg) since we found X above 0.155 m, we can solve for k, to obtain

k = 660 N/m.

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