B) For the situation described above, what is the work done on the particle by t

Question

B) For the situation described above, what is the work done on the particle by the force from x = 1.00 m to  x = 18.00 m.

C) For the situation described above, what is the velocity of the particle when it reaches x = 18.00 m.

D) For the situation described above, what is the impulse that the particle gives to you as it moves from x = 1.00 m until it reaches x = 18.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative.

Shown below is a graph of the 1-dimensional, net force as a function of position, F), that you exert on a particle of mass 12 kg -10 -12 -14 0 2 4 6 1 12 1416 18 x (meters)

Explanation / Answer

Part B)   work done on the particle by the force from x = 1.00 m to  x = 18.00 m

=>    work done on the particle = Area under graph from x = 1.00 m to  x = 18.00 m

                                                  = (1/2 * (5.571 - 1) * 8) - (1/2 * (13 -5.571) * 13) - ((18-13) * 13)

                                                   =   - 95 J

Part C)    Here,   - 95   + 1/2 * 12 * 6.2902   =    1/2 * 12 * v2

             =>     v = 4.871 m/sec               -------------------------->   velocity of the particle when it reaches x = 18.00 m

Part A)     impulse that you give to the particle   = change in momentum of particle

                                                                           =   12 * (4.871 - 6.290)

                                                                           = - 17.028 N.sec

Part D)     impulse that the particle gives to you =   17.028 N.sec

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