Video: Calculating the Motion of an Object Using Newton's Laws in Vector Form

An object is acted on by three simultaneous forces. 𝐅₁ = (βˆ’3.00𝐒 + 2.00𝐣) N, 𝐅₂ = (6.00𝐒 βˆ’ 4.00𝐣) N, and 𝐅₃ = (2.00𝐒 + 5.00𝐣) N. The object experiences acceleration of 4.23 m/sΒ². The object is initially at rest. Find the mass of the object. Find the speed of the object after 5.00 s. Find the components of the velocity of the object after 5.00 s.

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Video Transcript

An object is acted on by three simultaneous forces. 𝐅 one equals negative 3.00𝐒 plus 2.00𝐣 newtons, 𝐅 two equals 6.00𝐒 minus 4.00𝐣 newtons, and 𝐅 three equals 2.00𝐒 plus 5.00𝐣 newtons. The object experiences acceleration of 4.23 meters per second squared. The object is initially at rest. Find the mass of the object. Find the speed of the object after 5.00 seconds. Find the components of the velocity of the object after 5.00 seconds.

Given the three forces 𝐅 one, 𝐅 two, and 𝐅 three acting on an object initially at rest, which result in an acceleration of 4.23 meters per second squared, we want to solve first for the mass of the object. We can label that π‘š. We also want to solve for the speed the object attains after 5.00 seconds. We’ll call this speed 𝑠. And finally, we want to solve for the velocity components of the object after that same amount of time. We’ll call these components 𝑣 sub π‘₯ and 𝑣 sub 𝑦.

To start on our solution, we can draw a sketch of the forces acting on this object. If we sketch in the three forces acting on our object on a pair of axes in units of newtons, we see that we can add these together algebraically to solve for the net force acting on the object. Writing out 𝐹 one, 𝐹 two, and 𝐹 three by their component parts, when we add them together, we find a net resultant force of 5.00𝑖 plus 3.00𝑗 newtons.

Now that we know the net force, we can use Newton’s second law, which says this net force equals the object’s mass times its acceleration to solve for object mass π‘š. By the second law, object mass is equal to the magnitude of the net force acting on it divided by its acceleration π‘Ž. We solve for the magnitude of the net force by squaring each of its two components, adding the results together, and taking the square root of that sum.

Dividing that magnitude by our acceleration, given as 4.23 meters per second squared, we find that object mass, to three significant figures, is 1.38 kilograms. That’s the mass of the object under the influence of these three forces.

Next, we want to solve for the speed of our object after a time of 5.00 seconds has elapsed. Since the acceleration, in this case, is a constant value across that time, this tells us that the kinematic equations of motion apply to this object’s motion. We look through these different equations seeking a match between what we’re trying to solve for, final speed, and the information we already know.

Since we’re told that time, which we can call 𝑑, is 5.00 seconds, the first equation written down is a match for what we want to solve for. Written in terms of our variables, this equation is 𝑠 is equal to the initial speed of the object plus its acceleration times time. We know that because it started at rest, its initial speed is zero. So, its final speed is simply π‘Ž times 𝑑, or 4.23 meters per second squared times 5.00 seconds. This is equal to 21.2 meters per second. That’s the object’s final speed after 5.00 seconds under this acceleration.

Lastly, we want to solve for the components of the velocity of the object at the same time 𝑑 equals 5.00 seconds. To start on that solution, let’s draw the net force acting on our object in on our graph. Our net force, the sum of 𝐹 one, 𝐹 two, and 𝐹 three, is pointed at some angle, we can call it πœƒ, from the positive π‘₯-axis. This angle is important because the direction of our net force by Newton’s second law is the same as the direction of the resulting acceleration. And since our particle started from rest, that means its velocity will also be in the direction of force since it’s in the direction of acceleration.

Here’s what we know so far about our final velocity at 𝑑 equals 5.00 seconds. We know, first of all, that its magnitude is equal to 𝑠, which we’ve solved for. And second, we know that this velocity has an π‘₯- and a 𝑦-component. But we just don’t know what those components are. However, if we solve for the direction of the net force, then we can figure out how the final speed magnitude divides up over the two components of velocity.

So, looking at our diagram, if we draw in the π‘₯- and 𝑦-components of the net force and recognize we know their numerical values from solving for them earlier, we can write that the tangent of our angle πœƒ is equal to 3.00 newtons divided by 5.00 newtons. Taking the arctangent of both sides, we now have an expression for the angle πœƒ. And when we solve for it, we find it’s roughly equal to 31 degrees.

Now, that we know the angle πœƒ, we can solve for the components of velocity based on the final magnitude of that velocity, which is speed. We can write 𝑣 sub π‘₯ as the magnitude of the velocity, which is 𝑠, times the cos of πœƒ and 𝑣 sub 𝑦 as 𝑠 times the sin of πœƒ. Plugging in for these values when we solve for 𝑣 sub π‘₯ and 𝑣 sub 𝑦, we find they’re equal, to three significant figures, to 18.1 meters per second and 10.9 meters per second, respectively. In component form then, our final velocity is 18.1𝑖 plus 10.9𝑗 meters per second.

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