What is the total work done on the object?

In summary: 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
  • #1
MIA6
233
0
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.
 
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  • #2
MIA6 said:
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.

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...)
 
  • #3
MIA6 said:
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.

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?
 
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  • #4
MIA6 said:
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?

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...)
 
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  • #5
dynamicsolo said:
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...)

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

1. What is considered as "work" in the context of an object?

In physics, work is defined as the transfer of energy to or from an object by means of a force acting on the object. It is typically measured in joules (J) and is represented by the equation W = F*d, where F is the force applied and d is the displacement of the object in the direction of the force.

2. How is the total work done on an object calculated?

The total work done on an object is calculated by multiplying the force applied to the object by the distance the object moves in the direction of the force. This can be represented by the equation W = F*d. If multiple forces are applied, the total work done is the sum of the work done by each individual force.

3. Is the total work done on an object always positive?

No, the total work done on an object can be either positive or negative. If the force and displacement are in the same direction, the work done is positive. However, if the force and displacement are in opposite directions, the work done is negative. This is because work is a measure of the energy transferred to or from an object, and it can be gained or lost depending on the direction of the force.

4. Does the mass of an object affect the total work done on it?

Yes, the mass of an object does affect the total work done on it. This is because the force required to move an object a certain distance is directly proportional to its mass. Therefore, a larger mass will require more work to be done on it to achieve the same displacement as a smaller mass.

5. How does the angle between the force and displacement affect the total work done on an object?

The angle between the force and displacement affects the total work done on an object through the use of the cosine function. The total work done is equal to the magnitude of the force multiplied by the displacement multiplied by the cosine of the angle between them. This means that if the angle is 0 degrees (force and displacement are in the same direction), the total work done is maximized, and if the angle is 90 degrees (force and displacement are perpendicular), the total work done is zero.

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