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!

 

[Nate810]

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.

 

[ogb p]

I can provide a response to this scenario using the principles of force and work. First, let us determine the magnitude of force F when the crate is in its final position. To do this, we can use the equation F = mg, where m is the mass of the crate and g is the acceleration due to gravity. Substituting the given values, we get F = (230 kg)(9.8 m/s^2) = 2254 N.

Next, let us calculate the total work done on the crate during its displacement. Work is defined as the product of force and displacement, so we can use the equation W = Fd. Substituting the values, we get W = (2254 N)(4.00 m) = 9016 J.

Now, let us determine the work done by the gravitational force on the crate. Since the crate is hanging from a rope, the gravitational force is acting downwards, while the displacement is horizontal. Therefore, the work done by the gravitational force is zero, as the force and displacement are perpendicular to each other.

The work done by the pull on the crate from the rope can be calculated using the same equation, W = Fd. However, in this case, the force and displacement are in the same direction, so the work done by the rope is W = (2254 N)(12.0 m) = 27,048 J.

Since the crate is motionless before and after its displacement, the net work done on it is zero. This means that the work done by the force F must be equal to the negative of the work done by the other two forces, i.e., W(F) = -W(gravitational force) - W(rope pull). Substituting the values, we get W(F) = -9016 J - 27,048 J = -36,064 J.

Finally, the work done by your force F is not equal to the product of the horizontal displacement and the answer to A (2254 N) because the force is not acting in the same direction as the displacement. In this case, the force is acting at an angle to the horizontal direction, so only the component of the force in the horizontal direction is doing work. This is why we use the equation W = Fd cosθ, where θ is the angle between the force and the displacement. Therefore, the work done by your force F is W