Forces & Kinematics: Study with Video Lessons, Practice Problems & Examples

In solving one-dimensional motion problems involving forces, utilize the equation $F=ma$ alongside the equations of uniformly accelerated motion (UAM). Key variables include initial velocity, final velocity, displacement, time, and acceleration. For example, a 20 kg block accelerating from rest to 30 m/s in 6 seconds requires calculating acceleration as the link between force and motion. This results in an applied force of 100 N, demonstrating the integration of force and motion concepts effectively.

Solving Motion Problems with Forces

Hey guys. In earlier chapters, we saw how to solve motion problems using these four equations of motion right here. In more recent videos, we've started to see how to solve forces problems. Well, you're going to start to see some problems that now combine forces and motion. So I'm going to show you in this video how we solve 1-dimensional motion problems with forces. And really, it's just using a combination of equations that we've already seen before. So let's go ahead and check this out. But first, I want to recap what we've already seen in earlier chapters. Remember that your five motion variables are v_initial, v_final, delta x, delta t, and a. So we have these five motion variables. And as long as you know 3 out of those 5, you can pick one of these 3 or 4 equations to solve a problem. And in more recent videos, we started working with forces problems. Remember that there's only 3 variables in forces problems. That's f, m, and a, and there's only one equation to use. We use f=ma, which means that you need 2 out of those 3 knowns in order to solve a problem. But now we're going to start seeing problems that combine the two. You're going to start seeing combinations of force variables and motion variables. A diagram of that might look something like this where you have a mass that's being pulled or pushed by some force. It's going to accelerate, but you also have some initial velocity, final velocity, delta x or delta t. So to solve these problems, we're just going to use a combination of these equations. We're just going to use a combination of f=ma and our UAM or motion equations that we've already gotten super familiar with. So what I want you to do is look at these sets of variables here, and you'll notice that there's one that's common between them. You'll notice the a variable, the acceleration is the one that is shared between both sets. So that's how we're going to solve these kinds of problems. To show you how that works, I'm just going to go right into this example down here. So if we look at this example, we've got this 20 kilogram block. So that's a mass. We know that the mass is 20. It's on a frictionless surface. It's being pushed, but it's accelerating. It's accelerating to 30 meters per second from rest. What that means is that we have an initial velocity that's 0, but our final velocity is 30. And then we've got the 6 seconds over here. That's a time. So that's delta t. What we're going to do is we're going to calculate the magnitude of the applied force that's pushing the block. But this is a forces problem. So the way we solve forces problems is first we have to draw a free body diagram. Let's go ahead and do that. So we're going to draw the free body diagram like this. This is a friction of the surface. And the first thing we have to do is check for any weight force. Right? There's always a weight force unless we're told that there's no weight force. So there's a weight right here. Then we look for any applied or tension forces. Remember that this block is being pushed, so we can just assume that that is an applied force like this. So that's my f_a. There are no cables or tension or ropes or anything like that, so there's no tension. And the next thing we will look for is if two surfaces are in contact. Well, there are because this block is on some surface like this. So there's a normal force. That's the reaction to the surface push. And then we check for any friction last. Remember, this is a frictionless surface, so there is no friction or anything like that. So this is the free-body diagram. So the second step is now we're just going to write f = ma. So we're going to go ahead and do that. So I've got f = ma over here, the net force. Now in this particular case here, we're calculating the magnitude of the applied force. So what we're really looking for is looking for f_a. And because this is the only force that's actually pushing this block to the right, then that's the only force that we're going to look at in our f = ma equation. So we've got f_a = ma over here. So to solve this, I need the mass and the acceleration. So we've got f_a =, then I've got this as 20, that's the mass, and then I need the acceleration. The problem is that I actually don't know what that acceleration is. I don't know what this a variable is. So I need it in order to solve for my applied force. So if I can solve for this a, then I can figure out my f, my applied force. So how do I figure out this acceleration? Remember that what we said here was that acceleration was the shared variable between these two sets of variables. It's in f = ma, but it's also inside all of my motion equations and my motion variables. So what I can use is I can use one of these motion equations to solve for the acceleration. So really just going to use the same process that we've used before. So I need 3 out of 5 variables. I've got my v_naught, v_final, delta x, and delta t. So what happens is, I've got 3 out of those variables here and I can just pick one equation that's going to give me my acceleration. So really what happens here is that this acceleration variable a is the link between your force and your motion problems. If you ever get stuck using f = ma, then you can always use your UAM equations to solve and also vice versa. If you get stuck using UAM equations, you can use f = ma. So what I've got here is I've got 3 out of 5 variables, which means I can pick one of my 5 equations or 4 equations. That's going to be the first one, which is that v_final = v_initial + at. So I've got my v_final is 30, my v_initial is 0, then I've got a times 6. And if you go ahead and work this out, you're going to get 30 over 6 which equals the acceleration, which is 5. So now what we can do is just plug this variable back inside of this equation over here. So I've got my applied force is 20 times 5, and my F_a is equal to 100 newtons. And that's the answer. There's a 100 newton force that's pushing the block causing it to accelerate. That's it for this one, guys. Let me know if you have any questions.

An 800-kg car is traveling along a horizontal road directly towards a cliff. The driver notices and brakes, resulting in a 5,000-N net force slowing the car down. If the car's initial speed was 20 m/s and the car stops just before going over the cliff, how far away was the car from the cliff when the driver hit the brakes?

• A. 1250m
• B. 0.03m
• C. 64m
• D. 32m

Liftoff!

Alright, guys. Let's check out this next one here. We have a 1,000 kilogram rocket that is accelerating upwards due to some thrust. So I've got this rocket like this. I know the mass is 1,000, and I got a couple of forces on it. During the first 20 seconds of its motion, we know the weight force or the force of gravity downwards, this is my w, is going to equal 10,000. We also have a couple of other forces like the force of thrust. That's basically like an applied force. So what I'm gonna do is I'm gonna call this f_t for thrust. I also have a force of air resistance that's acting downwards because this rocket is gonna start accelerating. We're gonna start moving upwards like this. So what happens is the air resistance, I'm gonna call this f_air, is going to point downwards, and I know this is 5,000. I also know this thrust is 25,000. Alright? So I've got all these forces. There's no normals or frictions or anything like that. What I wanna do is figure out the rocket's velocity after 20 seconds. We know that the initial velocity is going to be 0 because it's going to start from rest, but then later on, it's going to have some final velocity. So this final velocity is actually what I'm looking for here. So if I wanna figure out v_final, I'm gonna have to list out all of my other kinematics variables, right, like my initial velocity. I also want the Δy, and then I have the acceleration and time. So I'll need all of these things. The Δy, I don't know what that is. I know that v₀ is 0, starts from rest. The acceleration, I don't know either and I also, I know the time is 20 seconds. Right? So this is going to be my 20 seconds here. So we have here that the 2 out of 5 variables. Right? We have 2 out of 5, which means we can't actually use an equation to solve for v_final yet. We're gonna need another one of our variables. Just like we have done before, when we get stuck, we're gonna try to use this, we're gonna try to figure out this a by using f=ma. So we're gonna use f=ma to figure out acceleration and we're now just gonna add all of our forces in the vertical direction. Right? All of our forces act only along the y-axis. So what happens is this rocket is traveling upwards, so I'm just gonna choose the upwards direction to be positive. And so that means that our f_t is positive when we expand out our Σf. f_air is going to be downward so it gets a negative. And then our weight force is also going to be negative as well. So this equals mass times acceleration. Remember, I wanna figure out this a here so I can plug it back into this, into my kinematics variables and then pick an equation. Alright? So now I just replace all the values that I know. This is 25,000 Then I've got this f_air is 5,000, and then I've got minus this 10,000 from the weight. And this equals mass which is 1,000 times acceleration. Alright, so if you work all this out what you're gonna get is 10,000. So you're gonna get 10,000 over here is equal to 1,000 times a. So that just means that our acceleration is 10 meters per second squared. Alright? So our acceleration is 10, which means we can now have enough information to figure out a kinematics equation, right? So we now we have 3 out of 5, and now we're gonna go ahead and just pick out one of your equations. So I'm gonna go over here. If I wanna find out the final velocity, the final velocity is basically going to be the equation that ignores my delta x. Right? So I'm gonna ignore my Δy over here, and that's just gonna be equation number 1. So I've got equation number 1, which is v_final equals v_initial plus at so if we want v_final this is just gonna be 0 plus now I've got an acceleration of 10 and then I've got in a time of 20. So this is basically just gonna be 200 meters per second. Alright? So look through your answer choices and it's answer choice a. That's it for this one, guys.

Here’s what students ask on this topic:

What are the equations of uniformly accelerated motion (UAM)?

The equations of uniformly accelerated motion (UAM) are essential for solving kinematics problems. They are:

1. $v_f=v_i+at$

2. $v_{f}2=v_{i}2+\; 2a\mathrm{\Delta x}$

3. $\mathrm{\Delta x}=v_{i}t+\; 0.5at2$

4. $\mathrm{\Delta x}=v_{f}t-\; 0.5at2$

5. $\mathrm{\Delta x}=\; 0.5(v_i+v_f)t$

How do you solve a problem involving both forces and motion?

To solve a problem involving both forces and motion, follow these steps:

1. Identify the known variables for both forces (F, m, a) and motion (v_i, v_f, Δx, Δt, a).

2. Draw a free-body diagram to visualize the forces acting on the object.

3. Use $F\; =\; ma$ to relate the forces to the acceleration.

4. If acceleration (a) is unknown, use one of the UAM equations to solve for it.

5. Substitute the acceleration back into $F\; =\; ma$ to find the force.

By combining these steps, you can effectively solve problems that integrate both force and motion concepts.

What is the significance of the acceleration variable in solving combined force and motion problems?

The acceleration variable (a) is crucial in solving combined force and motion problems because it serves as the link between the two sets of equations. In force problems, acceleration is used in $F\; =\; ma$. In motion problems, acceleration appears in the UAM equations. By determining acceleration from motion variables (v_i, v_f, Δx, Δt), you can use it to find the force using $F\; =\; ma$. This interconnectedness allows for a comprehensive approach to solving problems that involve both forces and motion.

How do you draw a free-body diagram for a forces problem?

To draw a free-body diagram for a forces problem, follow these steps:

1. Represent the object as a simple shape (e.g., a box).

2. Draw arrows to represent all forces acting on the object. Label each force (e.g., F_a for applied force, F_g for gravitational force).

3. Include the weight force (F_g = mg) acting downward.

4. Add normal force (F_N) acting perpendicular to the surface.

5. Include any applied or tension forces.

6. If applicable, add frictional forces.

This diagram helps visualize the forces and is essential for applying $F\; =\; ma$ to solve the problem.

What is the relationship between force, mass, and acceleration?

The relationship between force, mass, and acceleration is described by Newton's Second Law of Motion, which states that the net force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). This is expressed as:

$F\; =\; ma$

This equation indicates that for a given mass, the acceleration of an object is directly proportional to the net force applied. Conversely, for a given force, the acceleration is inversely proportional to the mass.