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.
Our first order of business (following our problem-solving guide process) is restating our problem and capturing the critical information that will help us set the foundation for the rest of the solution.
Below we address the variables the problem gives (mass, angle of the incline, distance the crate travels and the coefficient of friction).
We’ll also state the unknowns the problem asks us to find (work done by friction and and work done by gravity).
Now we’ll throw out our guess (a little memory trick that helps frame the problem).
Here, all I did was try 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 pulled a number out of thin air, I’m not worried about whether it’s actually right or not.
Now that we have a solid foundation of what we think our answer could be, we can actually try to work through the problem by re-drawing our problem statement and labeling our variables.
Diagram and Variables
Here we’ve included the mass of our crate (m), the angle of the surface in which the crate lies (𝜽) and the distance the crate travels (d). The coefficient of kinetic friction (μk) is not labeled but is also included in our problem setup.
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 and to the center of the crate, the force of gravity is pointing straight downwards towards the Earth, the normal force points opposite that of gravity and perpendicular to the surface the crate lies on, and the frictional force occurs between the crate and it’s surface and also points opposite the direction of motion.
We also established our x and y axis so that our y is perpendicular and upwards from the surface that the box slides along, and our x is parallel to the surface, facing downwards, as this is the direction of motion.
Next, we created a second FBD that gives us a slightly different perspective, where all of our forces have been placed on the x and y axis.
The force due to gravity (the only one not parallel or perpendicular to the slope surface) has been separated by using trigonometry, and the other acting forces have just simply been translated 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 unknown forces we may need to solve to problem.
Essentially, work is just force multiplied by distance, and this is the key equation we’ll use to determine the work done by friction as well as the work done by gravity.
In our equations we’ve also included the general formula for force, as well as the specific formula for the frictional force – two additional pieces of information we’ll need for our solution.
The work done by friction is observed by multiplying the force along our assigned x-axis (frictional force) by the distance the crate travels on that same axis (variable d). As seen below, we can simply substitute our force of friction equation into our work equation for F to solve for work.
Work done by gravity can be found in a similar way. Now, 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 multiplied by acceleration (as seen above). However, since we were given only the distance the crate slides along the sloped surface we don’t know the distance it goes only downwards.
Using our knowledge of sines and cosines we can solve that though. Our height equation listed below will will tell us the vertical distance travelled, it is an arrangement of SOH (sine equals opposite over hypotenuse)
Knowing the steps to solve for work done by friction and gravity, along with the equations required 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 our distance the crate travels along the assigned x-axis we are good to go there. However, we don’t know the force.
Recall the force done by friction equation, frictional force equals 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. Once again recalling from our FBD we know the normal force is equal and opposite to the contact force caused by the object. We happened to solve for this in our FBD (m x g x cos(𝜽)).
Knowing all of our variables, we can plug in and solve, but there is a catch. Since our crate is sliding downwards and our frictional force is acting in the opposite direction (opposes motion) it is making negative progress against the direction of motion. This means the distance it travels is in the opposite direction from how friction is acting, so our work is negative.
Now we can “plug and chug.”
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 we want to first use our work equation but substitute height (h) for distance (d) because we are now finding the vertical distance (W = F x h). Like we discussed earlier the work done by gravity is the force times distance in the downward direction.
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). SInce the force is known 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 needed to obtain all known variables within our equation. We can plug our known force into our work equation along with our known distance and solve.
This time work is positive because the force done by gravity and the direction of motion for our crate are in the same direction. Finally, we can just plug in our variables and solve.
Our result is 5,987 J but 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 work 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!