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The figures below show systems containing a block initially held at rest on a frictionless surface. In each system, the block is attached to the end of a spring, Which is stretched to the right. The mass and spring constant are given for each system, as well as the distance the spring is initially stretched. When the mass is released, the spring will accelerate the block. A: 5N/m spring constant, 0.4m stretched, 1kg mass B: 5N/m spring constant, 0.2m stretched, 2kg mass C: 4N/m spring constant, 0.2m stretched, 5kg mass D: 1N/m spring constant, 0.5m stretched, 1kg mass E: 4N/m spring constant, 0.5m stretched, 4kg mass F: 1N/m spring constant, 0.5m stretched, 5kg mass Rank these systems on the basis of the magnitude of the inital acceleration of the blocks from greatest to least. Please show all equations used and explain your reasoining.

Respuesta :

Answer:

1) Case A: a = 2.0 m/s^2

2) Case B, D, and E: a = 0.5 m/s^2

3) Case C: a = 0.16 m/s^2

4) Case F: a = 0.1 m/s^2

Explanation:

Given:

- A: 5N/m spring constant, 0.4m stretched, 1kg mass

- 5N/m spring constant, 0.2m stretched, 2kg mass

- 4N/m spring constant, 0.2m stretched, 5kg mass

- 1N/m spring constant, 0.5m stretched, 1kg

- 4N/m spring constant, 0.5m stretched, 4kg mass

- 1N/m spring constant, 0.5m stretched, 5kg mass

Find:

- Rank these systems on the basis of the magnitude of the inital acceleration of the blocks from greatest to least.

Solution:

- Develop a free body diagram of the block. There is only one force acting on the block the moment it is released i.e springs restoring force.

                                       F_sp = k*x

- Apply Newton's second law of motion:

                                       F_net = m*a

                                        k*x = m*a

                                        a = k*x / m

- After obtaining the general expression for acceleration we will compute for each case:

Case A: 5N/m spring constant, 0.4m stretched, 1kg mass

                                 a = (5)*(0.4) / 1 = 2.0 m/s^2

Case B: 5N/m spring constant, 0.2m stretched, 2kg mass

                                  a = (5)*(0.2) / 2 = 0.5 m/s^2

Case C: 4N/m spring constant, 0.2m stretched, 5kg mass

                                  a = (4)*(0.2) / 5 = 0.16 m/s^2

Case D: 1N/m spring constant, 0.5m stretched, 1kg mass

                                  a = (1)*(0.5) / 1 = 0.5 m/s^2

Case E: 4N/m spring constant, 0.5m stretched, 4kg mass

                                 a = (4)*(0.5) / 4 = 0.5 m/s^2

Case F: 1N/m spring constant, 0.5m stretched, 5kg mass  

                                 a = (1)*(0.5) / 5 = 0.1 m/s^2

Rank:

1) Case A: a = 2.0 m/s^2

2) Case B, D, and E: a = 0.5 m/s^2

3) Case C: a = 0.16 m/s^2

4) Case F: a = 0.1 m/s^2

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