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Work and Energy Question?

A girl pushes a 3.0 x 10^2 N box along the floor for a distance of 4.0 m. The coefficient of kinetic friction is 0.33. She then lifts the box straight up 1.2 m. What is the total amount of work she has done

Ok so the fg=fn but not this case because the girl is pushing down but yet idk by what force

so i do 300*4 to get work done by the girl =1200

then i find the work done friction by doing 300*0.33 =99*4 =396

1200-396=804

then i use Ep=mgh assuming mg=300 i do 300*1.2 =360 i do 804+360=1164

but the answer is 760J

2 Answers

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  • 8 years ago
    Favorite Answer

    The only work that is being done during the pushing is the work against friction. Therefore your work done for this part is your normal force (1,200J) times your coefficient because this is the portion of the normal force that is actually causing resistance. Because think about it, if the object were on near frictionless ice you could easily push it no matter it's weight, however if it is on a rug of something with a coefficient like .33 it is much more difficult. Anyway, so you would multiply, 1,200J x (1/3) = 400. It is always better to use a fraction if possible, because you know .33 = 1/3. Then you did the next part right, find the work done vs. gravity. You got 360J if you add the 400J for the previous part, boom you get 760J. You can see how close you were, just remember when you're pushing something, you're doing work against friction.

  • 8 years ago

    You are fundamentally wrong in one place, you say the weight of the box but don't mention the FORCE APPLIED in moving the box when she pushes it.As you know, Work done=Force X displacement, so it is wrong. You have only multiplied displacement with the wt. of box, which is not applicable when u push something horizontally, as wt. acts vertically and has no component along horizontal direction. So check if applied force is given, and try again.

    Source(s): I'm all into physics.
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