# Application of Vectors: Work (calc 3)

• ek530n
In summary, The person is seeking help with a calculus problem involving a box being pushed up a horizontal ramp. They have tried using the formula W=F(dot)D=|F||D|cos a with the given numbers, but the answer is incorrect. Another person suggests considering the vectors and using the formula W=F(dot)D instead. The person seeking help clarifies that the problem neglects gravity and units are not required. There is some confusion about the use of vectors and the calculator being in the correct mode.

#### ek530n

Hey all,

I'm a bit stuck on a problem on my online homework for my calc 3 class, hopefully someone can help me out.

Suppose that you push with a horizontal force on a box, to push it up a horizontal ramp, as shown in https://instruct2.math.lsa.umich.edu/webwork2_course_files/ma215u05/tmp/gif/2-prob3-pimages/sfig13-3-3.gif [Broken]

If your force is F=22 lbs., the ramp angle a(alpha)=17 degrees above the horizontal and you push the object a distance L=6 feet how much work is done on the box?

Thanks

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What ideas do you have...? I bet the problem is not that difficult.

Daniel.

Well I've tried the plug n' chug method with the numbers give:

W=F(dot)D=|F||D|cos a
W=(22 lbs.)(6 ft.)(cos 17)
W=126.232 ~ 126

But this answer is incorrect, unless the answer registered in the system is incorrect.

I don't know about your weird American system of units, but the formula is the right one.

Daniel.

I got the same answer. It looks fine to me. Did you put in the correct units?

Force is a vector.

It has X and Y components.

Also remember about Mass x Gravity.

Yes but the formula W=|F||D|cos a can be used instead of W=F(dot)D in order to avoid vectors and instead use magnitudes only.

Also gravity is neglected in this particular problem and units are not required.

ek530n said:
Yes but the formula W=|F||D|cos a can be used instead of W=F(dot)D in order to avoid vectors and instead use magnitudes only.

Also gravity is neglected in this particular problem and units are not required.

Eh? Thats odd. Ive never heard of a problem neglecting gravity before.

also isnt (dot) multiplication.

NOTE: to a helper who is reading this: I really need help with my topci before 2 hours, please!

Scirel said:
Eh? Thats odd. Ive never heard of a problem neglecting gravity before.

Is that an attempt to mock me for my choice of words? If you don't have anything relevant to say don't post under this thread.

Wow! Take it easy! I wasnt trying to make fun of you. Geez..

I literally have never seen a problem like that before. How can that be taken as an insult?

anyway, there is one more thing to try.

Are you sure your calculator is in degrees/radians mode? (for whatever the problem calls for)

ek530n said:
Well I've tried the plug n' chug method with the numbers give:

W=F(dot)D=|F||D|cos a
W=(22 lbs.)(6 ft.)(cos 17)
W=126.232 ~ 126

But this answer is incorrect, unless the answer registered in the system is incorrect.
You are given the x-component of the force
namely
22 lbs.=(Force)(cos(17 degrees))
You want work
work=force*distance
In other words your cos(17 degrees) should have be sec(17 degrees) or 1/cos(17 degrees).

## What is the definition of work in terms of vectors?

In physics, work is defined as the product of the magnitude of a force and the displacement of the object in the direction of the force. In terms of vectors, work is calculated by taking the dot product of the force vector and the displacement vector.

## How is the work done by a constant force calculated?

The work done by a constant force is calculated by multiplying the magnitude of the force by the displacement of the object in the direction of the force.

## How is the work done by a variable force calculated?

To calculate the work done by a variable force, the force must be broken down into small segments. The work done by each segment is then calculated using the dot product of the force vector and the displacement vector of that segment. The total work is the sum of all the individual segment works.

## What is the relationship between work and energy?

Work and energy are closely related concepts. Work is the energy transferred to an object when a force is applied to it. Similarly, the work done by an object is the energy it transfers to another object. This is known as the work-energy theorem.

## Can work be negative?

Yes, work can be negative. This occurs when the force and displacement vectors are in opposite directions, resulting in a negative dot product. Negative work means that the object is losing energy, while positive work means the object is gaining energy.