Force applied at an angle to a box?

AI Thread Summary
A net force of 4N is applied to a 6kg box moving 3m to the right, resulting in 12J of work done, calculated using W=Fdcos(0). For the second question, when the force is applied at a 30-degree angle, the work done can still be calculated using the same formula, but with the angle affecting the cosine component. The user expresses confusion about needing acceleration or velocities for the second part, but the focus should remain on the angle's impact on work. Ultimately, the key takeaway is that the angle alters the calculation, but the fundamental approach remains consistent.
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Homework Statement


A net force of 4N to the right is applied to a box of 6kg while moving it a distance of 3m to the right.
(1)How much work is done on the box?
(2)What if the force is applied at a 30 degree angle above horizontal?


Homework Equations


(1) W=Fdcos(angle)
(2) F=ma
(3) W=Kf-Ki=((1/2)mv^2)-((1/2)mv^2)

The Attempt at a Solution


For the first question, to find the work done on the box, I plugged numbers into equation (1).
W=(4N)(3m)cos(0)
Cos(0)=1, W=12J

I don't even know where to begin for question (2). It seems to me that, in order to caclulate Force, I need the acceleration to plug into equation 2, or I need velocities to calculate work in equation (3). I have neither, and I can't think of any other applicable equations. I assumed friction was negligable since it was not mentioned.
 
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why wouldn't plugging numbers in equation 2 work here as well? You already have the force.
 
Okay, I feel crazy. I got work and force mixed up and what the questions were asking mixed up. I feel dumb.

So, equation (1) is used for both questions. The force is the same, the distance is the same, the only thing that changes is the angle.

Thanks.
 
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