# Solve Work with Friction: Mass, μ & W Given

• Alex126
In summary, the problem involves finding the distance an object with a known mass and friction coefficient moves when a known work is applied. The solution involves using the equations W=F*s and F= -Friction Force + Force_that_pushes_the_object, with the assumption that the object moves in a straight line and begins and ends at the same speed. However, the problem is ambiguous and requires making assumptions, so it is important to clearly state those assumptions when solving the problem.
Alex126
Hi. The problem is really simple I think, but I feel like the text is wrong even though the result seems to be correct.

1. Homework Statement

An object of known mass m (71.5 Kg) is pushed on a [flat] surface with a known friction coefficient of μ (0.272). How much does the object move with a work W (642 J) ?

## Homework Equations

W = F*s
Friction force = Normal force * μ

## The Attempt at a Solution

The way I solved it, and the result seems to be correct at 3.36 meters, is like this.

Friction force = m*g*μ (since it's a flat surface, the Normal force = Weight force) = 191 N

W = F*s
642 = 191 * s => s = 642/191 = 3.36

The problem is, I feel like it's missing a piece of the puzzle, and more specifically the Force F that pushes the object in the first place (called F? in the drawing).

What I mean is, the Work should be given by F*s, but "F" here should be the total Forces along the "movement axis" of the object...right? So I thought that the F in W = F*s in this case should have been:

F = -Friction Force + Force_that_pushes_the_object

I figured I'd try solving the exercise assuming that F = Friction Force, and it turns out that the result is in fact correct. However, I feel like saying that the result "matches the result of the workbook" is more accurate than saying that it's "correct", because I really believe that the F should be = Force_push - Friction Force, and not F = Friction Force.

So, in my opinion, the text should have given the additional information on how much F? is worth, and then the solution should have been:

s = Work / (F? - Friction)

Am I wrong?

You are expected to read the problem statement as specifying the work done by the unknown force which you call F. You are also expected to assume that the object moves in a straight line (or that you measure distance along its path), that it begins and ends the scenario with the same speed, that air resistance is negligible and that this is done on Earth where the acceleration of gravity is 9.8 meters/sec2.

If the object starts and ends at rest and if the 642J is taken as the net work done by all forces on the object then this would violate the work-energy theorem. So that interpretation cannot be correct.

In the interest of brevity, almost all physics problems are stated without spelling out all the details. Often this means that they become ambiguous and sometimes hard to read. With practice, you can develop the skill to see which interpretations make sense and test the subject matter you have been learning.

One key rule applies: If you have to make assumptions about the problem to solve it, state those assumptions clearly and then work the problem. Even if the assumptions you make turn out not to be the ones intended, this doing this should at least earn you partial credit.

To be precise, the question should have been worded "how much work is the friction force doing on the mass, as reckoned from the laboratory frame of reference?" We don't know how much work the force F is doing on the mass, because we don't know if the mass is accelerating. The friction force F is doing positive work on the mass, while the frictional force is doing negative work (the displacement is opposite to the direction of the frictional force). If the force F just matches the friction force, then the work of the two forces cancel, and the kinetic energy of the mass stays constant. If the force F is larger than the friction force, there is net positive work done on the mass, and its kinetic energy (velocity) increases.

Alex126 said:
F = -Friction Force + Force_that_pushes_the_object
Correct (provided you drop that negative sign). But the push required to keep a body moving at a steady velocity is zero Newtons, in the absence of friction.

Ok, thanks all for the answers.

Just a little doubt.
Chestermiller said:
The friction force F is doing positive work on the mass, while the frictional force is doing negative work (the displacement is opposite to the direction of the frictional force).
You meant "the force F?" there?

Alex126 said:
Ok, thanks all for the answers.

Just a little doubt.You meant "the force F?" there?
Yes. Sorry.

## 1. What is the formula for calculating work with friction?

The formula for calculating work with friction is W = F * d * cosθ, where W represents work, F represents the force applied, d represents the displacement, and θ represents the angle between the force and displacement vectors.

## 2. How does mass affect the amount of work done with friction?

The mass of an object does not directly affect the amount of work done with friction. However, a heavier object may require more force to overcome friction and thus result in more work being done.

## 3. What is the coefficient of friction and how does it impact work done with friction?

The coefficient of friction (μ) is a measure of the amount of friction between two surfaces. It represents the ratio of the force required to move an object to the normal force between the two surfaces. A higher coefficient of friction means that more force is needed to overcome friction, resulting in more work being done.

## 4. Can work be negative when dealing with friction?

Yes, work can be negative when dealing with friction. This occurs when the force applied is in the opposite direction of the displacement, resulting in negative work being done.

## 5. How can the work done with friction be minimized?

The work done with friction can be minimized by reducing the force or the coefficient of friction. This can be achieved by using lubricants, polishing surfaces, or using materials with lower coefficients of friction.

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