Calculating Forces and Work in Crate Displacement

[BnJ]

A 230 kg crate hangs from the end of a rope length L=12.0 m. You push horizontally on the crate with a varying force F to move it distance d=4.00m to the side.
A: What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are B:the total work done on it, C:the work done by the gravitational force on the crate, and D: the work done by the pull on the crate from the rope? E: Knowing that the crate is motionless before and after its displacement, use the answers to B, C, and D to fine the work your force F does on the crate. F: Why is the work of your force not equal to the product of the horizontal displacement and the answer to A?

I am clueless...please help!

 

[member 553083]

A: The magnitude of F is equal to the weight of the crate, W=mg=230 kg x 9.8 m/s^2 = 2,254 N. B: The total work done on the crate is equal to the product of the force and displacement, W=Fd= 2,254 N x 4.00 m = 9,016 J.C: The work done by the gravitational force on the crate is equal to the product of the weight of the crate and the vertical displacement, W=mgΔh= 230 kg x 9.8 m/s^2 x 12.0 m = 27,312 J.D: The work done by the pull on the crate from the rope is equal to the product of the force from the rope and the displacement, W=Fd= (2,254 N - mg)d = 2,254 N x 4.00 m - 230 kg x 9.8 m/s^2 x 12.0 m = -18,296 J.E: The work your force F does on the crate is equal to the total work done on the crate minus the work done by the gravitational force and the work done by the pull on the crate from the rope, W=Wtotal-Wgrav-Wrope = 9,016 J - 27,312 J - (-18,296 J) = 54,624 J. F: The work of your force is not equal to the product of the horizontal displacement and the answer to A because the crate also experiences a vertical displacement. Additionally, the force of gravity and the pull from the rope are also acting on the crate, so these must be taken into account to determine the work your force F does on the crate.