A 16.0 kg block (m1) is on a 30 degree incline. A second block m2 (12.0 kg) is a

Question

A 16.0 kg block (m1) is on a 30 degree incline. A second block m2 (12.0 kg) is attached to it by a massless cable stretched over a massless, frictionless pulley. When released, both of the blocks are at a height of 1.25 m. The static friction coefficient of the ramp is 0.400 and the kinetic friction coefficient is 0.320.

1) Show that the blocks initially do not move.

An external force F is applied to block m1 continually in a horizontal direction.

2) What value of the external force F will just start the blocks moving?

3) What is the acceleration of the blocks just before m2 hits the ground?

4) What is the tension in the cable just before m2 hits the ground?

5) What is the acceleration of m1 after m2 hits the ground?

6) What is the acceleration of m1 as it starts back down the ramp?

7) What is the tension in the cable when m1 stops?

8) When m1 was initially moving up the incline, what value of the external force F would have caused the acceleration constraint to change such that a_1a_2?

Here is a link to a visual representation:

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Explanation / Answer

1)     initially, Force of gravity on m1 = 16 * 9.8 * sin30 = 78.4 N

                      Force of friction on m1 = 0.4 * 16 * 9.8 * cos30 = 54.31 N

=>    T - 78.4 - 54.31   = 16a

=> T   -   132.71 = 16a

Also,   12a = 117.6 - T

=>   acceleration is negative so, blocks initially do not move .

2)     Here, Fcos30 + 12*9.8 = 16 * 9.8 * sin30 + 0.4 * 16 * 9.8 * cos30

   => F   =   17.45 N

3)     here, acceleration of the blocks just before m2 hits the ground = 0.471 m/sec2

4)    Tension in cable =   117.6 - 12*0.471

                                    =   112 N

5)    acceleration of m1 after m2 hits the ground = 9.8 * sin30 = 4.9 m/sec2

6)    Here, acceleration of m1 as it starts back down the ramp =   15.11/28   =   0.539 m/sec2

7)    tension in the cable when m1 stops   =   117.6 + 12*0.539

                                                                  = 124.068 N

8)    value of the external force F would have caused the acceleration constraint = >

   Here, F   = (16 * 9.8 * sin30 + 0.4 * 16 * 9.8 * cos30 )/cos30

        => F   =   153.24 N

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