In this article (and video above), we take a deep dive into how to solve a two-block system problem using the work-energy principle, straight from the Dynamics section of the FE Exam. We explore how forces like friction, gravity, and motion all interact in a real-world scenario involving an inclined plane and a pulley. By breaking down the steps and applying core principles, you’ll learn how to calculate the velocity of a block after it descends a given distance helping you confidently prepare for the FE Exam.

**Question:**

In this problem, we have two blocks connected by a massless rope over a frictionless pulley. Block A, weighing 25 kg, rests on a rough 30° inclined plane with a kinetic friction coefficient of 0.15, while Block B, weighing 35 kg, hangs vertically. The system is released from rest and our goal is to determine the velocity of Block B after it descends 8 meters.

**Problem C****ontext****:**

Before diving into the solution, let’s review the problem context, starting with a clear understanding of the work-energy principle. This principle tells us that the total work done by all forces acting on a particle is equal to the change in the particle’s kinetic energy. In other words, if we can account for all the forces involved — whether they’re conservative forces like gravity or non-conservative ones like friction — we can figure out how the energy in a system changes.

Now, let’s break this down even further by defining some key concepts. Firstly, work is defined as the product of a force and the displacement of an object in the direction of that force:

**U = F ∙ d**

This concept will be critical when dealing with forces like friction or gravity that influence the blocks in our system.

Next is kinetic energy. This is defined as the energy possessed by an object due to its motion. As a block in our system speeds up, its kinetic energy increases, and we can calculate its kinetic energy at a specific point in time using the familiar formula:

**T = 1/2 mv ^{2}**

Lastly, we have potential energy, which is the energy stored by an object based on its position or condition. For example, when a block is hanging at a height, it has gravitational potential energy that can be converted into kinetic energy as it falls:

**V = mgh** (gravitational potential energy)

When we combine these three concepts, we arrive at the work-energy equation:

**T _{1} + V_{1} + U_{(1→2)} = T_{2} + V_{2}**

This equation tells us that the initial kinetic energy T_{1}, the initial potential energy V_{1}, and any external work done by non-conservative forces U_{(1→2)}, must be equal to the sum of the final kinetic energy T_{2} and final potential energy V_{2}. In other words, this equation represents the conservation of energy. While energy can transform between kinetic and potential forms, it cannot be created or destroyed. This principle will allow us to solve our system involving the two blocks.

**Step 1 – Evaluate T**_{1}:

_{1}:

Let’s start applying this principle to our problem. We can successfully evaluate this scenario by substituting for the unknowns in our equation, starting with T_{1}. Since the system starts from rest, both blocks A and B have an initial velocity of zero. The kinetic energy of each block is given by 1/2 mv^{2}, so with no initial velocity, the initial kinetic energy of both blocks will be zero. To calculate the initial system kinetic energy T_{1}, we can simply add the respective kinetic energies of both blocks, but since they’re zero in this case, we know that our entire T_{1} term reduces to zero in our work energy equation.

1. Initial kinetic energy of block A

2. Initial kinetic energy of block B

3. Initial system kinetic energy

**Step 2 – Evaluate V**_{1}:

_{1}:

Next, let’s consider the system’s initial potential energy, V_{1}. Block B is hanging vertically, and its potential energy is determined by its height. The higher it is, the more gravitational potential energy it has. Similarly, Block A, on the inclined plane, also has potential energy based on its height relative to the ground. But how can we specify the height of each block relative to the ground if we have no idea how far the ground is situated from our system? Well, when referring to each object’s height, we are referring to a relative quantity. This means that we have to define a reference point for each object before we can assign a quantity to its relative height.

We know that Block B will travel a vertical distance of 8 meters after which the system will arrive at its second state, so to simplify this problem, we can define our point of reference for block B at this final location.

1. Initial potential energy of block B

This leaves block B at a height of 8 metres ABOVE our reference location, which allows us to obtain an expression for its initial potential energy.

We can use a similar approach to determine the relative height of Block A. Since Blocks A and B are connected by a rope, we know that when Block B moves 8 meters downward, Block A must also move 8 meters, but along the incline. However, for gravitational potential energy, we only care about Block A’s VERTICAL displacement and to calculate this, we’ll use trigonometry.

Since the incline is at an angle of 30°, the vertical height Block A moves is given by h_{B} sin(θ) or = 8 sin(30°). As with block B, we can use this height as our reference point. We now know what the vertical distance travelled by block A is between states 1 and 2. But remember, Block A starts BELOW its reference point. So, its initial height relative to the reference point is NEGATIVE, which means its initial potential energy is also negative. We now find the potential energy of block A as:

**V _{A1} = m_{A}∙g∙h_{A} = (25)(9.81)(-8 sin(30°))**

And finally, now that we have the initial potential energy of both blocks—Block B’s positive and Block A’s negative—we can add them together to obtain a formula for the total initial potential energy of the system. By substituting the respective values, we find that the system has an initial potential energy of 1765.8 J.

**Step 3 – Evaluate U**_{1→2}:

_{1→2}:

We can now move on to the work being done on the blocks as they move from state 1 to state 2. Since Block A is moving along a rough incline, we need to account for the work done by friction, a non-conservative force. The work done by friction can be calculated by multiplying the kinetic friction force, F_{k}, and the displacement of Block A along the incline, here referred to as d:

**U _{friction} = -F_{k} d**

The negative sign is there because friction always opposes the motion, so it does negative work on the system.

To calculate the friction force acting on Block A, we need to analyze the forces in its free-body diagram. First, we know that the friction force is given by the equation:

**F _{k} = μ_{k} N**

where μ_{k} is the coefficient of kinetic friction, and N is the normal force. This friction force always opposes the motion of the object. Since Block A is moving up the incline, friction will act in the opposite direction—down the incline. But to get the normal force, we need to break down the gravitational force acting on Block A. The weight of Block A pulls straight down with a force equal to the mass of A times gravity. Now, since Block A is on an inclined plane, this weight can be split into two components: one that acts parallel to the incline, pulling Block A downward along the slope, and one that acts perpendicular to the surface, pressing it against the plane.

The perpendicular component is what we’re interested in, because that’s the force that determines the normal force. Using some basic trigonometry as before, we find that this perpendicular component is equal to:

**N = F _{⊥} = m_{A}∙g∙sin(θ) = (25)(9.81) cos(30°)**

Finally, we can now calculate the friction force using our equation:

**F _{k} = μ_{k} N**

And when substituting the respective values we arrive at a friction force of 31.85 N.

And with this friction force known, we can finally solve for the WORK done by friction. By substituting F_{k} as 31.85 N and the distance travelled as 8 metres, we find that the work done by friction is equal to -254.87 J:

**U _{friction} = -F_{k} d = -(31.85)(8) = -254.87 J**

The key point here is that we use the full 8 meters as the distance, because friction is continuously applied as Block A moves up the incline.

This allows us to move on to the right side of our equation, where we need to solve for the total kinetic and potential energy of our system during its second state. Let’s start with the kinetic energy, T_{2}. The total kinetic energy is the sum of the kinetic energies of both blocks, A and B. We also know that the formula for kinetic energy is 1/2 mv^{2}, and that we need to apply this for each block. But since Block A and Block B are connected by a rope, they move together. This means that both blocks share the same final velocity after Block B descends the 8 meters. And remember, this final velocity is the quantity we’re interested in. We treat the final velocity of both blocks as a single variable. But we’ll keep it as an unknown for now. This leaves us with an expression for T_{2}, defined as:

**T _{2} = 1/2 (25) v^{2} + 1/2 (35) v^{2}**

Now, as soon as we’ve defined our final variable V_{2}, we’ll be ready to solve for this variable v, or otherwise, block A and B’s final velocity.

To calculate the system’s final potential energy, we need to go back to the reference heights we set earlier. We chose these references to be at the blocks’ final positions—meaning when Block B has descended 8 meters and Block A has moved up the incline. By doing this, we’ve strategically set both blocks to have a height of zero relative to these reference points in their final positions. Since gravitational potential energy depends on an object’s height above a reference point, this means the potential energy of both blocks in state 2 is zero. So, the total final potential energy of the system is also zero, meaning we can go ahead and cross V_{2} off from our work-energy equation.

And now that we’ve defined all of our variables in our equation, from T_{1} to V_{2}, we can solve for the final velocity of the blocks.

By substituting the known values for potential energy, kinetic energy, and the work done by friction, we’re left with a single unknown- the final velocity, v. From here, we just rearrange the equation to make v the subject and solve for it.

**The answer is A!**

And that completes our problem! By browsing the options given to us, we can see that the final answer is equal to A.

**Final Remarks:**

Before we wrap up, let’s go over a few key takeaways from this problem. First, always be strategic when choosing reference heights for potential energy—here, we set them at the final positions of the blocks to simplify our equation, but remember that you could choose them at any location that makes the most sense to you. The key, is just to stay consistent with your choice. Second, don’t forget to account for friction when dealing with rough surfaces. In this problem, friction worked against Block A, so we had to include its negative impact on the system’s energy. Thirdly, since Block A and Block B are connected by a rope, their motions are linked, meaning they must have the same velocity at all times. This is a crucial detail that’s easy to overlook, but it’s essential for solving the problem correctly. Always remember, when two objects are connected like this, their velocities will be equal, even if they’re moving in different directions or on different paths. Finally, the best way to sharpen your skills is by practicing with a wide range of problems. In this case, mastering the underlying principles and how to apply them is the key to success. This will ensure you can confidently tackle any problem, even if it’s presented in a slightly different way.

**About Matthew Douglas**

Matthew currently serves as a content creator and host of The Engineering Project Management Podcast. A civil engineer by trade, Matthew has developed a passion for construction and stormwater management by way of maintenance and rehabilitation services. Matthew has also had experience working with private consulting firms and public agencies, and has even held a role of an educator. As such, he loves to lead, build, mentor, and help those in need.

Most recently, during his time working for the public sector, he has taken the role of Public Works Operations Manager. He led quite a few public infrastructure rehabilitation projects and implemented new asset management technologies at a very young age. It is here that the passion for “fixing what’s broken” developed.

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I hope you found this article helpful. In upcoming articles, I will solve some more PE exam practice problems and answer other questions from our subscribers. Pass the PE Exam videos will publish weekly, so be sure to click the subscribe button so you don’t miss something that could make a substantial difference in your exam result.

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I’ll see you next week… on Pass the FE Exam

Anthony Fasano, P.E.

Engineering Management Institute

Author of Engineer Your Own Success

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