conservative forces

MIA6

1. An object has a mass of 8.0 kilograms. A 2.-newton force displaces the object a distance of 3.0 meters to the east, and then 4.0 meters to the north. What is the total work done on the object?
1)10 J 2) 14 J 3) 28 J 4) 56 J
I found the resultant of the two distances, which is 5 m, I used 5 m * 2 N=10 J, but it’s not the answer.

2. A 20-newton block is at rest at the bottom of a frictionless incline, since I can’t show you the diagram, I will describe it. Supposed a right triangle, the vertical height is 3 m, and the base is 4.0 m. Then the hypotenuse is the incline and the block is at the bottom. The question is: How much work must be done against gravity to move the block to the top of the incline?

I know that conservative forces are forces for which the work done does not depend on the path taken but only on the initial and final positions. But here I Know I should use 3 m * 20N=60 J. but the thing is it starts from the bottom which is the bottom of the incline, and ends at the top, then it should be the length of the hypotenuse? But if it’s 3 m, then it should have started from left end of the base, not the right end of the base which is the bottom of the incline.

Thanks a lot for help.

dynamicsolo

The work done is the product of the path length times the component of the force along that path. Note that for this problem, the force first pushes the mass 3 m. east, then turns and pushes it 4 m. north. So the direction of the force is not constant and the work will not be related to the net displacement (though, naturally, they've provided a choice for anyone who read the problem that way -- as I probably would have the first time...)

buffordboy23

First, W = F dot d = Fd cos(theta), which is applicable to situations of constant force (F) and angle (theta) between the displacement (d) and force. Is the direction of force in the same direction for both the east and north displacements? If the force pushed the object all over the place and then back to the original starting point, is the net work done zero?

dynamicsolo

The work done by gravity is the force of gravity (here, the weight of the block) times the component of the path in the direction gravity acts (this is the alternate interpretation of the vector dot product F · delta_x to that used in problem 1). So the work done by gravity will be mg (downward) · 3 m. (upward) · cos 180º = 20 N · 3 m. · (-1) = -60 J. The work done against gravity is thus done by an external agent (like, say, you!), and will be just the negative of this result, or +60 J.

The horizontal displacement of the block by 4 m. does not require work to be done against gravity, since gravity only acts vertically. In terms of the dot product, the work done by (or against) gravity in the horizontal direction would be 20 N · 4 m. · cos 90º = 0 .

This raises an interesting practical question: if the work done against gravity when moving things horizontally is zero, why do I get so tired from moving furniture around? The issue there is that to hold and carry objects with mass at a constant height above the floor, you must hold your muscles tensed to support the object's weight the entire time. So, even though you are doing no work against gravity, you are still using your own internal energy to manage the task. (This is the drawback of being an animate support for a weight. For an inanimate support, like a hook in the ceiling or a nail in the wall, the energy is simply provided from stretching the interatomic bonds in the structure of the metal a bit...)

MIA6

ook. I get it. so it's 2N*3m + 2N *4m=14J. Thanks.