Force applied at an angle to a box?

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SUMMARY

The discussion focuses on calculating work done on a box when a net force of 4N is applied at different angles. For a horizontal force, the work done is calculated using the formula W = Fdcos(angle), resulting in 12J when the angle is 0 degrees. When the force is applied at a 30-degree angle above the horizontal, the same formula applies, but the cosine of the angle must be considered, leading to a different calculation for work. The participant expresses confusion about the relationship between force, acceleration, and work, but ultimately recognizes that the angle is the only variable that changes in the work equation.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with the work-energy principle (W=Kf-Ki)
  • Knowledge of trigonometric functions, specifically cosine
  • Basic physics concepts related to force and motion
NEXT STEPS
  • Learn how to calculate work done with forces applied at angles using W=Fdcos(angle)
  • Study the implications of friction on work calculations in physics
  • Explore the relationship between force, mass, and acceleration in Newton's laws
  • Investigate the work-energy theorem and its applications in different scenarios
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of work calculations involving forces at angles.

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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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