Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed a

Question

Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring constant 650 Nm. The other end of the spring is attached to a wall. The block is pushed toward the wall until the spring has been compressed a distance x. The block is released and falls 0.80m vertically from the edge of the table, and striking a target on the floor that is a horizontal distance of 1.2m from the edge of the table. Air resistance is negligible. A) calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor, B) calculate the speed of the block as it leaves the table, C) Calculate the distance x the spring is compressed

Explanation / Answer

a vi = 0 a = 9.81 m/s^2 d = 0.8 m t = ???? d = vi*t + 1/2 a*t^2 0.8 = 0 + 1/2 * 9.81 * t^2 t ^2 = 0.163 t = 0.404 s Part B ===== The speed of the block (as it leaves the table) is all horizontal. That can be determined by using d = v * t There is no acceleration in the horizontal direction. d_h = 1.2 m t = 0.404 seconds v = d/t = 1.2 / 0.404 = 2.98 m/s Part C ==== The only way I can think to do this (and I'm not sure it's right, but there is no other hidden fact I can see) is to assume that the spring is providing enough force to move the 4 kg mass (turned into a a weight). F_mass = 4 * 9.81 = 39.24 N. In other words the spring has to provide 39.24 N F = 39.4N k = 650 N / m x = ???? 39.4 = 650 * x x = 0.060 m = 6.0 cm

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