Question

Two horizontal forces F1 and F2, are acting on a box. Fi points to the right. 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 = +3.9 N and the mass of the box is 3.2 kg. Find the magnitude and direction of F2 when the acceleration of the box is (a) +4.8 m/s

, (b) -4.8 m/s
, and (c) 0 m/s
.

Answer 1

Using Newton's second law, we find that when the box accelerates at +4.8 m/s², the force F2 is 11.46 N to the right, at -4.8 m/s², F2 is 19.26 N to the left, and at 0 m/s², F2 is 3.9 N to the left.

Step-by-step explanation:

The question asks to determine the magnitude and direction of a force F2 that acts on a box which is subject to another force F1 and accelerates along the x-axis without friction. We use Newton's second law, Fnet = ma, to solve for F2. The mass m of the box is 3.2 kg, and force F1 is given as +3.9 N.

1. When the acceleration of the box is +4.8 m/s²:
   
2. When the acceleration of the box is -4.8 m/s²:
   
3. When the acceleration of the box is 0 m/s²:
   

SUMUP all the final answers as points:

• Magnitude of F2 when a = +4.8 m/s² is 11.46 N to the right.
• Magnitude of F2 when a = -4.8 m/s² is 19.26 N to the left.
• Magnitude of F2 when a = 0 m/s² is 3.9 N to the left.