Work Done By Friction (Physics 1 Problem Solution)

Having a hard time calculating the work done by friction?

Turns out all it takes is a few FBDs, a little trig, and understanding the work equation.

In this post and video, we go through a solution to a work done by friction (and work done by gravity) physics problem involving a man, an incline, and an unfortunate situation with a heavy crate.

Work Done By Friction Problem Statement

A 250-kg crate slides 5.2 m down a 28-degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.

The effective coefficient of kinetic friction between the crate and the surface is 0.40.

Calculate the work done by friction and the work done by gravity.

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Problem Restatement

Our first order of business (following our problem-solving guide process) is to restate our problem and capture the critical information we need for the rest of the solution.

Bonus: Download the full version of this Work Done by Friction Problem Solution (with annotations) you can take with you.

Below we address the variables the problem gives (mass, angle of the incline, distance the crate travels and the coefficient of friction).

We also state the unknowns the problem asks us to find (work done by friction and work done by gravity).

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Now for a quick guess (a little memory trick that helps frame the problem).

Here, I’m just trying to approximate the work done by friction by multiplying mass, gravity, distance, and the coefficient of friction.

To find work done by gravity I took mass x gravity x distance.

But regardless, even if I just pull a number out of thin air, I’m not worried about whether it’s actually right or not.

Instead, a guess anchors the solution to the problem around your initial intuition, which creates a stronger reference point in your memory.

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Now that we have an idea of what we think our answer could be, let’s start actually solving.

We’ll re-draw the diagram in the problem statement and label our variables.

Diagram and Variables

Here we include the mass of our crate (m), the angle of the surface on which the crate lies (𝜽) and the distance the crate travels (d).

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We also include the coefficient of kinetic friction (μk) in our problem setup, even though it’s not labeled.

Next step, we’ll draw a free-body diagram to display how the forces are acting on the crate as it slides down the slope.

Free Body Diagram

The initial free body diagram shows the actual forces and their angle of action on the crate.

The man pushes parallel to the slope acting on the center of the crate.

The force of gravity points down towards the Earth’s center.

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The normal force is a reaction in the opposite direction of gravity, perpendicular to the surface the crate lies on.

And the frictional force occurs between the crate and its surface, and points opposite the direction of motion.

We also establish our x and y-axes.

Our y is perpendicular and upwards from the surface that the box slides along.

Our x is parallel to the surface, facing downwards because this is the direction of motion.

Next, we create a second FBD that gives us a slightly different perspective.

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We’ll place all of our forces along the x and y-axis we just established.

Then we separate the force due to gravity (the only one not parallel or perpendicular to the slope surface) using trigonometry.

Then we can simply translate the other acting forces so that they all act from the center of the crate.

By doing this, we can now use force balance equations in our solution to determine the value of the unknown forces we may need to solve the problem.


Essentially, work is just force multiplied by distance.

This is the key equation we’ll use to determine the work done by friction as well as the work done by gravity.

We’ll also include the general formula for force, as well as the specific formula for the frictional force. These are two additional pieces of information we’ll need for our solution.

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We can find the work done by friction by multiplying the force along our assigned x-axis (frictional force) by the distance the crate travels on that same axis (variable d).

We’ll determine work done by gravity in a similar way. However, instead of the distance and force along the assigned x, we need to find the distance and force created by gravity going straight down.

The force in this direction is mass times acceleration. But since we were given only the distance the crate slides along the sloped surface, we don’t know the distance it goes only downwards.

Don’t worry… we can solve that though. Remember your trig?

Our height equation listed below will tell us the vertical distance traveled. It is an arrangement of SOH (sine equals opposite over hypotenuse).

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Now we have our basic equations and variables ready. Plus, we know the general steps to solve for work done by friction and work done by gravity.

So now we can go ahead with re-arranging and finally, plugging in.


Work Done By Friction

First, we take our general definition of work, work equals force times distance (W = F x d).

Since we know the distance the crate travels along the assigned x-axis we are good to go there. However, we don’t know the force due to friction.

Thankfully we have an equation for that 🙂 The frictional force equals the coefficient of friction times the normal force (Ff =μk x N).

If we plug this in for F that leaves N as our only unknown. Let’s look back at our FBD.

We know the normal force is equal and opposite to the contact force caused by the object. And look at that: we happened to solve for this in our FBD (m x g x cos(𝜽)).

Now that we know all of our variables, we can plug in and solve right? Ah, but there is a catch.

Since our crate is sliding downwards and our frictional force is acting in the opposite direction it is opposing the movement of the crate.

This means our work due to friction is negative, and is assisting our fellow here in slowing down the runaway crate.

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Now finally, we can “plug and chug.”

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Plugging in we get -4,504 J for our final work done by friction. Since the problem gave two significant figures we round up to -4,500 J.

Work Done By Gravity

Once again, let’s use our work equation but substitute height (h) for distance (d) because we want the vertical distance.

Like we discussed earlier, the work done by gravity is the force times distance in the downward direction (W = F x h).

Now to find the force.

Recall our general force equation, force equals mass times acceleration (F = m x a). Here we know our mass of the object (250 kg) and the acceleration due to gravity (9.81 m/s2).

Okay, we’ve got the force. Now we can move on to height.

The problem does not give us the height so we are left to solve.

Recall our height equation above, height equals distance times sine of theta (h = d x sin(𝜽)).

This is all we need to obtain all the known variables within our equation.

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This time work is positive because the force done by gravity and the direction of motion for our crate one in the same. Finally, we can just plug in our variables and solve.

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Our result is 5,987 J. Rounding up to two sig figs we get 6,000 J as our final answer for work done by gravity.

And we’re done!

The beauty of this type of problem is that once you recognize how to calculate the work done by both of these forces, the rest of the solution is just using good old fundamentals. Good luck!

Bonus: Download the full version of this Work Done by Friction Problem Solution (with annotations) you can take with you.
{ 2 comments… add one }
  • Emmanuel Kaira June 25, 2019, 5:16 pm


  • Francisco Silva September 10, 2019, 2:39 pm

    But since our N is perpendicular to the x-axis and parallel to the y-axis, why do we have to use cos(a)? Our reference axis accounts for the angle of the surface.

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