# Need help with finding Work from given force and distance components

• yang09
In summary: So you need to use tan-1((-4)/(8)), and then adjust the angle as necessary (the "standard" angle is, I think, in the 4th quadrant, so you may need to add 360° to it).Hope that helps! :smile:
yang09

## Homework Statement

A force ~F = Fx ˆı+Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 8 N, Fy = −4 N, sx = 5 m, and sy = 1 m.
Find the work done by the force on the
particle.

## Homework Equations

x^2 + y^2 = z^2
Work = (Force)(Distance)(cos(theta))

## The Attempt at a Solution

I found the magnitude of Force by squaring each component
F = (8)^2 + (-4)^2 = 8.9442N
I found the magnitude of Distance by squaring each component
F = (5)^2 + (1)^2 = 5.099m
I then created a triangle to find theta, using distance for my triangle sides.
tan(theta) = 1/5 so theta = inverse tan(1/5). I then got a theta of 11.310.
I plugged my force, distance and theta into the Work formula but I'm not getting the right answer.
W = (8.9442N)(5.099)(cos(11.310))
W = 44.721 Joules

## Homework Statement

Find the angle between ~F and ~s.

## The Attempt at a Solution

Welcome to PF!

Hi yang09! Welcome to PF!
(have a theta: θ and try using the X2 and X2 tags just above the Reply box )
yang09 said:
Work = (Force)(Distance)(cos(theta))

Work also = Force "dot" Distance, which is the easiest way to do it if you have the coordinates.

So what is (8,-4)·(5,1) ?

Thanks for the tips but what do you mean by:
What is (8,-4)·(5,1) ?
Do you want me to multiply them together or what? Or are you asking me what they stand for because the 8 is the x-component of force, -4 is the y-component of force, 5 is the x-component of distance, and 1 is the y-component of distance.
I'm pretty sure my magnitude is right, but am I doing something wrong with my θ?

Hi yang09!
yang09 said:
Thanks for the tips but what do you mean by:
What is (8,-4)·(5,1) ?

I meant the "dot" product of two vectors (also called the "scalar product" or the "inner product"), where you multiply the components and then add.

… am I doing something wrong with my θ?

Yes, your θ is tan-1(1/5), but that is the angle between (5,1) and the x-axis: you need the angle between (5,1) and (8,-4).

## 1. What is work?

Work is a physical concept that measures the amount of energy transferred by a force acting on an object as it moves a certain distance.

## 2. How is work calculated?

Work is calculated by multiplying the magnitude of the force applied to an object by the distance the object moves in the direction of the force.

## 3. What are force and distance components?

Force and distance components are the individual parts of a force or distance vector that act in different directions. In order to find work, these components must be added together.

## 4. How do I find work from given force and distance components?

To find work from given force and distance components, you must first calculate the total force and distance by adding together the individual force and distance components. Then, plug these values into the work formula: work = force x distance.

## 5. What are some examples of work in everyday life?

Some examples of work in everyday life include pushing a shopping cart, lifting a backpack, and pedaling a bike. In each of these examples, a force is applied to an object and it moves a certain distance, resulting in work being done.

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