# Work: Kinetic Friction + Applied Resultant Force

• HumKinStudent
In summary, the coefficient of kinetic friction between a 20kg box and the floor is 0.40. To find the work done by a pulling force applied at a 37 degree angle above the horizontal, the force should be broken down into horizontal and vertical components. Since the acceleration is zero, the sum of all horizontal components must be zero and the sum of all vertical components must be zero.
HumKinStudent

## Homework Statement

The coefficient of kinetic friction between a 20kg box and the floor is 0.40 -> How much work does a pulling force applied 37 degrees above the horizontal do on the box in pulling it 8.0m across the floor at constant speed

## Homework Equations

F(kinetic)= F(normal) x U(kinetic)

W=Fxd

Fx=Fcos(theta)

## The Attempt at a Solution

I solved for kinetic friction, I'm just not 100% sure where to go next, a hint would be more helpful in the long run than the answer to the entire question... I was thinking of assuming that force in the x direction should be equal to F(kinetic) and then just multiplying f(kinetic) by the distance 8 metres... Nonetheless I feel like there is more to it...

HumKinStudent said:
I'm just not 100% sure where to go next, a hint would be more helpful in the long run than the answer to the entire question... I was thinking of assuming that force in the x direction should be equal to F(kinetic) and then just multiplying f(kinetic) by the distance 8 metres... Nonetheless I feel like there is more to it...

There cannot be a single force in the x-direction if the box is moving at constant speed. Do you see why?

kuruman said:
There cannot be a single force in the x-direction if the box is moving at constant speed. Do you see why?

if the box is moving at constant speed in the x direction then the force MUST be equal to the kinetic friction, correct?

Should I go back, make my sum of Forces in the y = 0 and recalculate the normal force by factoring in the y component of the force, instead of assuming the normal force is the opposite of the box's weight x gravity?

This will change my kinetic force of friction...

HumKinStudent said:
if the box is moving at constant speed in the x direction then the force MUST be equal to the kinetic friction, correct?

It is correct to say that, because the speed is constant, the component of the pulling force in the horizontal direction and the force of kinetic friction add to zero.

Should I go back, make my sum of Forces in the y = 0 and recalculate the normal force by factoring in the y component of the force, instead of assuming the normal force is the opposite of the box's weight x gravity?

This will change my kinetic force of friction...

Yes you should and consider the component of the pulling force perpendicular to the incline.

kuruman said:
It is correct to say that, because the speed is constant, the component of the pulling force in the horizontal direction and the force of kinetic friction add to zero.

Yes you should and consider the component of the pulling force perpendicular to the incline.

there is no incline in this question, it is on a flat surface, nonetheless, I should still break down the resultant force...

Yes, there is no incline, I got confused. You should break down the puling force into horizontal and vertical components. Since the acceleration is zero, the sum of all the horizontal components must be zero and the sum of all the vertical components must be zero.

## 1. What is kinetic friction?

Kinetic friction is the force that opposes the motion of an object when it is in contact with a surface and moving at a constant velocity.

## 2. How does kinetic friction affect the motion of an object?

Kinetic friction causes an object to slow down and eventually come to a stop if there is no applied force to counteract it.

## 3. What factors affect the amount of kinetic friction?

The amount of kinetic friction depends on the type of surfaces in contact, the roughness of the surfaces, and the normal force applied between the two surfaces.

## 4. How is kinetic friction different from static friction?

Kinetic friction occurs when an object is already in motion, while static friction occurs when an object is at rest and trying to be put into motion.

## 5. How do you calculate the applied resultant force needed to overcome kinetic friction?

The applied resultant force needed to overcome kinetic friction is equal to the coefficient of kinetic friction multiplied by the normal force between the two surfaces. This can be represented by the equation F = μkN, where F is the applied resultant force, μk is the coefficient of kinetic friction, and N is the normal force.

• Introductory Physics Homework Help
Replies
7
Views
558
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
29
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
994
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
363
• Introductory Physics Homework Help
Replies
56
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
974
• Introductory Physics Homework Help
Replies
10
Views
4K