# Physics

Two horizontal forces, F1-> and F2-> , are acting on a box, but only F1->is shown in the drawing. F2-> can point either to the right or to the left. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that F1-> = +8.0 N and the mass of the box is 4.7 kg. Find the magnitude and direction of F2-> when the acceleration of the box is (a) +6.7 m/s2, (b) -6.7 m/s2, and (c) 0 m /s2.

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1. You first have to find the acceleration of what is given. F1 has force of +8.0N, and box has mass of 4.7kg.
F = ma, so 8.0/4.7 is 1.7m/s2.

Now to view the other question, you'd use the same equation only this time, you need to use the newly acquired acceleration from above.
F2 = (4.7)(6.7-1.7) = 23.5N
F2 = (4.7)(-6.7-1.7) = -39.48N
F2 = (4.7)(0-1.7) = -7.99N

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posted by Megan

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