Work/Force question, can anyone explain?

  • Thread starter Predator
  • Start date
  • Tags
    Explain
In summary, the conversation discusses finding the work done on a mobile object by a force, its average power, and the angle between two vectors. The suggested method of using the Pythagorean theorem to find the magnitude of the force is incorrect and instead, the dot or scalar product should be used to find the work. The distance over which the force is applied is the magnitude of the vector found by taking the difference between the initial and final positions. Average power is calculated by dividing the average work by the time over which it is applied.
  • #1
Predator
14
0
1. A force F = (2.80 N)i + (6.80 N)j + (7.05 N)k acts on a 2.00 kg mobile object that moves from an initial position of di = (3.05 m)i - (2.00 m)j + (5.00 m)k to a final position of df = -(5.00 m)i + (4.05 m)j + (7.00 m)k in 4.00 s.



Homework Equations



W = Fd


The Attempt at a Solution



I thought I could do the pythagorean theorem for the Force...

square root (2.8^2 + 6.8^2 + 7.05^2), that would give the magnitude of the force.

Then I did the same thing, the magnitude of df - magnitude of di.

Then, (magnitude F) * (magnitude df - magnitude di) = W

But that is incorrect. I am using Web assign online and the answer I get from the previous method does not work. Can anyone explain?
 
Physics news on Phys.org
  • #2
Oh, I forgot the actual question haha.

(a) Find the work done on the object by the force in the 4.00 interval

(b) Find the average power due to the force during that interval.

(c) Find the angle between vectors. di and df.
 
  • #3
Here is a definition of work
http://hyperphysics.phy-astr.gsu.edu/hbase/wint.html#wi

One could find it with the magnitude of force applied over the distance.

But then one simply needs to apply the dot or scalar product

F [itex]\cdot[/itex] s, where s is the difference between the position vectors df-di.

Average power is just average work (equivalent to energy) divided by the time over which the work is applied.


magnitude of df - magnitude of di
does not work in finding the distance.

The distance over which the force is applied is the magnitude of the vector which is found by taking the difference between the position vectors df - di.
 

1. What is work?

Work is defined as the application of force over a distance. In simpler terms, it is the use of energy to move an object from one point to another.

2. What is force?

Force is a push or pull that is applied to an object. It is a vector quantity, meaning it has both magnitude and direction.

3. What is the relationship between work and force?

The amount of work done on an object is directly proportional to the amount of force applied and the distance over which the force is applied. This is known as the work-energy theorem.

4. Can you explain work in terms of physics?

In physics, work is defined as the product of force and displacement. Mathematically, it is expressed as W = F x d, where W is work, F is force, and d is displacement.

5. How is work measured?

Work is measured in joules (J) in the SI system. It can also be measured in foot-pounds (ft-lb) or calories (cal) in other systems of measurement.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
549
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
198
  • Introductory Physics Homework Help
Replies
4
Views
585
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
212
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top