A 0.140 kg block is suspended from a spring. When a small pebble of mass 30 g is

Question

A 0.140 kg block is suspended from a spring. When a small pebble of mass 30 g is placed on the block, the spring stretches an additional 6.2 cm. With the pebble on the block, the block oscillates with an amplitude of 12.0 cm. (Assume that the pebble is glued to the block.)
(a) What is the frequency of the motion?
_____Hz

(b) How long does the block take to travel from its lowest point to its highest point?
_____s

(c) What is the net force on the pebble when it is at a point of maximum upward displacement?
______N (downward)

Explanation / Answer

Given that mass of the block is M = 0.140 kg mass of the pebble is m = 30 g                                          = 30*10-3 kg                                          = 0.03 kg total mass is M   = 0.14 kg + 0.03 kg                              = 0.17 kg Length of the spring when pebble is placed is x = 0.062m a )       force F = kx               mg = kx      (since , force F = mg) therefore , spring constant   k = m g / x                               = (0.03 kg)(9.8 m/s^2) / (0.062 m)                               = 4.74 N /m frequency of motion is                f = 1/2 k / M                   = (1 /2) (4.74 N/m / 0.17 kg)                    = 0.84 Hz b )       Time period of the pebble block system is                              T = 1/ f                                  = 1/0.84 Hz                                  = 1.189 s time taken by the block to travel from lowest point to highest point                   t = T /2                         = 1.189 s / 2                         = 0.594 s c ) net force on the pebble when it is at a point of maximum upward displacement is                  F = mg                      = (0.03 kg)(9.8 m/s^2)                      = 0.294 N Given that mass of the block is M = 0.140 kg mass of the pebble is m = 30 g                                          = 30*10-3 kg                                          = 0.03 kg total mass is M   = 0.14 kg + 0.03 kg                              = 0.17 kg Length of the spring when pebble is placed is x = 0.062m a )       force F = kx               mg = kx      (since , force F = mg) therefore , spring constant   k = m g / x                               = (0.03 kg)(9.8 m/s^2) / (0.062 m)                               = 4.74 N /m frequency of motion is                f = 1/2 k / M                   = (1 /2) (4.74 N/m / 0.17 kg)                    = 0.84 Hz b )       Time period of the pebble block system is                              T = 1/ f                                  = 1/0.84 Hz                                  = 1.189 s time taken by the block to travel from lowest point to highest point                   t = T /2                         = 1.189 s / 2                         = 0.594 s c ) net force on the pebble when it is at a point of maximum upward displacement is                  F = mg                      = (0.03 kg)(9.8 m/s^2)                      = 0.294 N

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