A box of mass 3.1kg slides down a rough vertical wall. The gravitational force o

Question

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed of 2.5m/s , you start pushing on one edge of the box at a 45? angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23N , as shown in figure. (Figure 1) There is now a frictional force between the box and the wall of magnitude 13N . How fast is the box sliding 2.5s after you started pushing on it?

Assuming that the angle at which you push on the edge of the box is again 45?, with what magnitude of force Fp should you push if the box were to slide down the wall at a constant velocity? Note that, in general, the magnitude of the friction force will change if you change the magnitude of the pushing force. Thus, for this part, assume that the magnitude of the friction force is f=0.566Fp.

Fp = ?

Explanation / Answer

Without being able to see figure 1 we can only guess at the solution, but I can tell you what to do. The extra push must be resolved into horizontal and vertical components Fh and Fv . Since the angle is 45 deg. each will have the same magnitude: 23 sin(45) N , but we need the diagram to see whether the vertical component is up or down. The net downward vertical force is now F = mg - 13N +- Fv
once you have F you get a = F / m = F / 3.1kg and finish with your Vi = Vf + at

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