A 2.5-kg box, sliding on a rough horizontal surface, has a speed of 1.2 m/s when it makes contact with a spring (see the figure. the block comes to a momentary halt when the compression of the spring is 5.0 cm. the work done by the friction, from the instant the block makes contact with the spring until is comes to a momentary halt, is - 0.50 j. (a what is the spring constant of the spring?

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Ayanna Hale

18 December, 20:43

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Aydin Yates · Answer 1

E = 0.5 * m * v² = 0.5 * k * x² - 0.5

m = 2.5

v = 1.2

x = 0.05

solve for k (spring constant).