How Do You Calculate the Pulling Force on a Crate for Zero Net Work?

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To calculate the pulling force on a crate for zero net work, one must understand that the work done by the pulling force must equal the work done against kinetic friction. The crate, weighing 116 kg and pulled at a 35.2° angle, experiences a kinetic friction force due to the coefficient of friction of 0.212. The key is to set the horizontal component of the pulling force equal to the kinetic friction force, ensuring that they cancel each other out. This means that while the crate may be moving, no net work is done on it. Understanding this balance is essential for solving the problem correctly.
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Hello! I need a little help here. I have been trying to figure out this problem, but I just can't and I don't know where I am going wrong.

A 116 -kg crate is being pulled across a horizontal floor by a force P that makes an angle of 35.2 ° above the horizontal. The coefficient of kinetic friction is 0.212 . What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

now I know what W=FdCos(theta) and that fs=uk*F...but I just don't know where to go from there and I don't know what the end of the question means..."so that hte net work done by it and the kinetic frictional force is zero?"...does that mean I just set the net work and frictional force equal and solve for the force?
 
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Any moving object on a surface with friction will have kinetic friction.

Ah I am starting to see now. The question intends to ask "what should P be if the work done by you and the work done by kinetic friction cancel each other out?" So yeah you're right - almost. You can't set work equal to force because they are not of the same unit. Set your horizontal component of P equal to the kinetic friction experienced because if both forces cancel each other out, no work can be done in moving the box even if it's moving at a constant speed.
 
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Thanks! That's exactly what someone said when I asked them about it today and it worked.
 
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