The spring in FIGURE EX10.21a is compressed by 10 cm. It launches a block across a frictionless surface at 0.50 m/s. The two springs in Figure EX10.21b are identical to the spring of Figure EX10.21a. They are compressed by the same 10 cm and launch the same block. What is the block's speed now?

Question

Accepted Answer

Hey, everyone in this problem, we're told that elastic chords can be coupled to achieve a desired force constant. We have a single cord stretched by 200 millimeters like a slingshot launches a cube at 0.89 m per second through a frictionless bench. Three chords which are identical to the first one connected as shown in the diagram are used to launch the same cube at the same stretching of 200 millimeters. Whereas to determine the cube speed when launched from the combined quarts. OK. We're given four answer choices all in meters per second. Option A 1.55 option B 17.3 option C 6.93 and option D 0.516. We have our diagram here. It shows us that our cube is 2 kg. We have a sp spring constant of 40 noons per meter for our cord. And then we have this stretch of 200 millimeters and then we have the exact same thing. But with three of those identical cords in the second case. So what we have here is we have some spring potential energy due to this cord. We have some kinetic energy due to the motion of the block. So let's consider the conservation of mechanical energy and we have no friction. We're told that this is a frictionless bench. So we have no net external forces so we can go ahead and use our conservation of mechanical energy. Now recall that the conservation of mechanical energy tells us that the initial kinetic energy k, what's the initial potential energy you know, is equal to the final kinetic energy? KF what the final potential energy you have? Now, the initial case we're gonna say is when we are at maximum stretch. OK. So when this thing is stretched by 200 mil millimeters, and that cube is actually going to be at rest the final position. UF we're gonna consider when the cube is moving cord has returned to its resting position. So it's no longer stretched. OK. Now, when we talk about potential energy here, in this case, we're only talking about the spring potential energy. We don't have to worry about potential energy due to gravity because this is on a bench and that's not going to affect what's happening. OK. That block is not moving vertically. All right. So if we're initially at rest, then knot is going to be zero. And at the end, if this cord is no longer stretched and it's returned to its resting position, then the potential energy is going to be easier. And so what we have is all of this initial potential energy from our cores being converted into kinetic energy for the cube. Now recall that the potential energy is given by one half multiplied by K multiplied by X squared. And the kinetic energy is given by one half multiplied by M multiplied by V squared. Now thinking about our potential energy, OK, we just said that it's one half kx squared, but we have the potential energy from our three strings or three chords that we have to consider. OK. So we need to add all of those up because these chords are all identical. They have the same value of K, they're stretched to the same position. So they have the same X value. What we can do is just say that this is gonna be three multiplied by one half kx squared. OK. So we're adding that three times the same thing is just multiplying by three. And on the right hand side, our final kinetic energy. So one half MVF squared. All right. So a lot happened there. But what we did is we got to the equation three multiplied by one half KX squared is equal to one half mvs squared. Let's go ahead and fill out everything we know. So on the left we have three multiplied by one half multiplied by K, which is that spring constant, which we're told in the diagram, 40 noons per meter multiplied by X which is that stretch all squared. Now, the stretch is 200 millimeters. And we wanna convert this to our standard unit of meters. So we get to that final answer of meters per second that we want to go from millimeters to meters. We divide by 1000. And so we end up with 0.2 m all square. And on the right hand side, we have one half multiplied by the mass of our cube 2 kg multiplied by VF squared. If we simplify, on the left hand side, we have 2.4 Newton meters. And on the right hand side, we have 1 kg multiplied by VF squared, dividing by 1 kg. We get that VF squared is going to be equal to 2.4 meters squared per second squared. And when we take the square root, we get a speed of approximately 1.5492 m per second. And that is that final speed of the cube using the three ports. So let's go ahead and compare this to the answer choices that we were given. And we're gonna need to round a little bit. If we round to two decimal places, we can see that the answer we found corresponds with answer choice. A 1.55 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

The spring in FIGURE EX10.21a is compressed by 10 cm. It launches a block across a frictionless surface at 0.50 m/s. The two springs in Figure EX10.21b are identical to the spring of Figure EX10.21a. They are compressed by the same 10 cm and launch the same block. What is the block's speed now?

Verified Solution

Key Concepts

Hooke's Law

Conservation of Energy

Kinetic Energy