Two boxes on a horizontal surface, as shown in the figure above, are moving toge

Question

Two boxes on a horizontal surface, as shown in the figure above, are moving together under the influence of a horizontal force, F. The coefficient of kinetic friction between the boxes and the surface is 0.40. Use g = 10 m/s2. In case (a), a horizontal force F directed right is applied to the larger box, which has a mass of 8.00 kg. In case (b), the horizontal force F is instead applied to the smaller box, which has a mass of 2.00 kg. The horizontal force F has a magnitude of 100. N.

Let's say that the boxes are moving to the right, in the direction of the force F. What is the magnitude of the force exerted by the larger box on the smaller box in case a and case b?

Explanation / Answer

Where 100N is applied in the direction of motion, friction opposes the 100N force. We first find net force and acceleration: Fnet = 100 – Friction Fnet = 100 – (0.40)(8.00 + 2.00)(10) = 60N a = Fnet/m = 60/(8.00 + 2.00) = 6m/s/s to the right. Both boxes are accelerating in the direction of the 100N force at 6m/s/s to the right. (a) From the description in the question, it would seem in “case a” that the large box is to the left of the small box. The net force on the small box is equal to the force applied by the large box, less friction. Fnet = Fapp – Friction (2.00)(6) = Fapp – (0.40)(2.00)(10) Fapp=20N (b) From the description in the question, it would seem in “case b” that the large box is to the right of the small box. The net force on the small box is equal to the 100N force less the force applied by the large box, and less friction. Fnet = 100 – Fapp - Friction (2.00)(6) = 100 – Fapp – (0.40)(2.00)(10) Fapp=80N

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