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A box of mass 3.06 kg slides down a rough vertical wall. The gravitational force on the box is 30 N . When the box reaches a speed of 2.5 m/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 = 23 N , as shown in (Figure 1) . There is now a frictional force between the box and the wall of magnitude 13 N . How fast is the box sliding 2.6 s after you started pushing on it?

I've already found the box's speed vf at 2.6 s after you first started pushing on it. which was 3.1 m/s

But now I'm being asked the following question And I need to know how to set it up and solve it showing all the steps. I've tried many ways put I'm just missing something

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

This was my question and the right answer and feedback I got was this:

N

the feedback given was:

"Your results make sense. If you push with a force of magnitude 23 N , as described in Part C, the box will continue to speed up because the y component of the net force, and therefore the y component of the box's acceleration, remains negative (i.e., it points downward). In this case, the effect of the pushing force is simply to reduce the magnitude of the box's acceleration. Then, to reduce the box's acceleration further to zero, you need to push with a force of magnitude Fp>23N , as you have just calculated. Note that if you pushed even harder, the acceleration will become positive, causing the box to slow down and possibly come to a stop. "

Can someone please show me how they got Fp = 24 N would be great.

N

the feedback given was:

Explanation / Answer

When you are pusing up with force Fp
lets balance the forces in vertical direction.
In upward direction there is friction force (0.566 fp ) and Fp*cos 45
In downward direction there is mg (30 N)
Balancing forces in vertcal deirection.
Fp *cos 45 + friction force = m*g
Fp *0.7071 + 0.566 Fp = 30
Fp = 30 / (0.7071 + 0.566)
=23.56 N
Hope it clear how you got Fp=24 N

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