Calculate Work Friction in Figure 8-16 | Walker2 8.P.002

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The discussion focuses on calculating the work done by friction on a 2.1 kg box sliding along a floor, considering three paths with a coefficient of kinetic friction of 0.21. The initial calculation suggests that the work done by friction is -12.98 joules, assuming all paths have the same displacement of 3 meters. However, participants debate the impact of vertical movement on friction, questioning whether work done in vertical directions should be considered since the normal force in those directions is zero. The consensus is that only horizontal movement contributes to work done by friction, leading to confusion about the correct approach to the problem. Ultimately, the key takeaway is that work done by friction should be calculated based solely on horizontal displacement.
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Figure 8-16: http://www.webassign.net/walker/08-16.gif

12. [Walker2 8.P.002.] Calculate the work done by friction as a 2.1 kg box is slid along a floor from point A to point B in Figure 8-16 along paths 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.21.

Since work can have a negative number, I'm assuming that all 3 answers will be the same, since the displacement is 3 meters. Path 1 travels 4 - 1 = 3 meters, Path 1 travels 2 + 1 = 3 meters, and Path 3 travels 3 meters. The vertical travels will not be affected by friction.

W = f * d
d = 3
f = friction
friction = mu * N (force Normal)
N = ma
N = 2.1 * 9.81
N = 20.601
friction = 0.21 * 20.601 = 4.32621
W = 4.32621 * 3
W = 12.98 j

but this is wrong, and so is -12.98.

Thanks in advance for your input :biggrin:
 
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Work is done in the direction of an applied force. In this case you can't neglect the vertical lines. Work is done on them too. You should add them up.
 
CartoonKid said:
... In this case you can't neglect the vertical lines. Work is done on them too...

But is work done on them by friction? All it asks for is the work done by friction. And friction is mu * N, but in the vertical direction doesn't N = 0? meaning that work done by friction would also be 0? I can see why there would be work done by gravity, but we have to ignore that. I think :confused:

In the previous problem, we had to figure out work done by gravity, but not friction and I got it right:
figure: http://www.webassign.net/walker/08-15.gif

[Walker2 8.P.001.] Calculate the work done by gravity as a 1.7 kg object is moved from point A to point B in Figure 8-15 along paths 1, 2, and 3.
The answers are -33j for all 3 parts.
 
Last edited:
This question is different, it is actually a plan. You look the movement of the object from above. The object is always on the floor.
 
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