# Finding work given mass and mew?

• m4ya
In summary: Yes, I understand vectors and dot products. So when you have two vectors, like the normal force and the force of friction, they add together but in a different way than if they were just two numbers?That's right. The dot product of two vectors always produces a vector with the same magnitude and direction as the original vectors. So the force of friction would add to the normal force, creating a vector that points in the opposite direction of the original normal force. But since the magnitude of the force of friction is smaller than the magnitude of the normal force, the vector would be smaller in magnitude.Sorry I forgot to mention the problem also states the surface is level, and the sleigh is moving with a constant acceleration.
m4ya

"someone is pulling a 200kg sleigh across a field at a constant velocity. If the coefficient of kinetic friction is .13, how much work do they do on the sleigh if they pull it 94 meters?"

I know I have to find force net and then use the formula for work (force x distance)

how would I go about finding force net? Force friction+Force normalizing (which would be mass x gravity?)

any guidance would be helpful asap. thanks!

The way the problem is written you can assume it's a level (horizontal) field.

m4ya said:

"someone is pulling a 200kg sleigh across a field at a constant velocity. If the coefficient of kinetic friction is .13, how much work do they do on the sleigh if they pull it 94 meters?"

I know I have to find force net and then use the formula for work (force x distance)

how would I go about finding force net? Force friction+Force normalizing (which would be mass x gravity?)

any guidance would be helpful asap. thanks!

You don't do any work on the normal force. It's perpendicular to the direction of displacement. Think vectors.

Sorry I forgot to mention the problem also states the surface is level, and the sleigh is moving with a constant acceleration. Would I begin by finding force friction and force normalizing?

You'll want the normal force first.

m4ya said:
Sorry I forgot to mention the problem also states the surface is level, and the sleigh is moving with a constant acceleration. Would I begin by finding force friction and force normalizing?

Yes, do that. But then tell me why only the frictional force is important for the work done.

So I multiplied

200kg x 9.8 =1960 (normal force)

1960 x .13 = 254.8 (Force friction)

Would these two be added together to equal force net?

or do I just use the force friction to find the work because the surface is level?

Are you lifting the sled, or dragging it?

pulling, there is no angle. It is a level surface

m4ya said:
So I multiplied

200kg x 9.8 =1960 (normal force)

1960 x .13 = 254.8 (Force friction)

Would these two be added together to equal force net?

or do I just use the force friction to find the work because the surface is level?

The forces would be added together as vectors. And the equation for the work is the dot product of the force with the displacement. You do know about vectors, right?

so work =force x distance

so 2214 N x 94 m =208,116...

this number seems to big. what am I doing wrong?

isn't it something along the lines of setting the two equal to 0?

m4ya said:
so work =force x distance

so 2214 N x 94 m =208,116...

this number seems to big. what am I doing wrong?

isn't it something along the lines of setting the two equal to 0?

You don't add forces pointing in different directions together to get a single number. You have to think about vectors. Each of those forces contributes to the work differently because the work depends on the direction of the force and the direction of the displacement. Not just the magnitude of each.

Last edited:
m4ya said:
or do I just use the force friction to find the work because the surface is level?

Yes! Why not include the normal force? Tell me why.

is it because the normal force is acting on the object both up and down? the surface is opposing 1960 N and the object is exerting 1960 N of force on it? so one is positive, one is negative, and they cancel? The i'd get the answer by multiplying the force friction by the distance?

m4ya said:
is it because the normal force is acting on the object both up and down? the surface is opposing 1960 N and the object is exerting 1960 N of force on it? so one is positive, one is negative, and they cancel? The i'd get the answer by multiplying the force friction by the distance?

That's actually a pretty good guess. But it isn't quite complete. If a force is acting on an object perpendicular to the direction of motion then no work is done. Work isn't as simple as magnitude of force*magnitude of distance. There are directions to think about. Do you about vectors and dot products?

## 1. What is the formula for finding work given mass and mew?

The formula for finding work given mass and mew is W = mew * m * g * d, where W represents work, mew represents the coefficient of friction, m represents the mass, g represents the acceleration due to gravity, and d represents the distance.

## 2. How do you calculate work when given mass and mew?

To calculate work when given mass and mew, you will need to use the formula W = mew * m * g * d. Simply plug in the values for mew, m, g, and d and then solve for W.

## 3. What units are used for mass and mew when calculating work?

Mass is typically measured in kilograms (kg) and the coefficient of friction, mew, is a unitless quantity. However, it is important to make sure that the units for mass and mew are consistent when calculating work.

## 4. Can you give an example of calculating work using mass and mew?

Sure, for example, if an object with a mass of 10 kg is pushed with a coefficient of friction of 0.5 over a distance of 10 meters, the work done would be W = (0.5)(10 kg)(9.8 m/s^2)(10 m) = 490 joules. Therefore, the work done would be 490 joules.

## 5. How does the coefficient of friction affect the amount of work done?

The coefficient of friction, mew, represents the amount of resistance an object will experience when moving over a surface. The higher the mew value, the greater the friction and the more work will be required to move the object over a distance. Therefore, the coefficient of friction directly affects the amount of work done when given mass and distance.

• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
28
Views
461
• Introductory Physics Homework Help
Replies
3
Views
761
• Introductory Physics Homework Help
Replies
7
Views
682
• Introductory Physics Homework Help
Replies
4
Views
758
• Introductory Physics Homework Help
Replies
56
Views
1K
• Introductory Physics Homework Help
Replies
57
Views
639
• Introductory Physics Homework Help
Replies
12
Views
723
• Introductory Physics Homework Help
Replies
13
Views
926
• Introductory Physics Homework Help
Replies
11
Views
1K