A 50 g rock is placed in a slingshot and the rubber band is stretched. The magnitude of the force of the rubber band on the rock is shown by the graph in FIGURE P9.56. The rubber band is stretched 30 cm and then released. What is the speed of the rock?

Question

Graph showing the force exerted by a rubber band on a rock as it stretches to 30 cm.

Accepted Answer

Welcome back, everyone. We are making observations about a 30 g ball or a 300.3 kg ball bearing. And we are told that it is launched from a stretched elastic cord and the final length of the stretched elastic cord before launch is 20 centimeters or 200. m. And we are tasked with finding what is going to be the final velocity of the ball upon launch here. Another piece of information that we are given to help us solve this problem is this force versus position of the elastic cord or the stretch um in reference to the ball bearing. And we know that the integral of this exact graph, our force, our elastic force with respect to X from the initial length of the chord to the final length of the chord is equal to work. Now, why am I drawing this comparison here? Well, we know that work is going to be the change in kinetic energy for elastic systems. This is one half times mass times our final velocity minus our initial velocity squared, but we have no initial velocity of the ball. So work on the left hand side will just be equal to one half times mass times our final velocity squared. Going back to our integral here. What is going to be our elastic force that we are going to be integrating here? Well, we can just treat the elastic cord as a spring and therefore our force will be equal to K times X. What we can do here, we can actually solve for this K constant using this point at the end of our graph, at the end of our graph, this will be our maximum force and our maximum length of the chord which we already know to be 0.0.2 meters. We won't plug that in right now because we're just trying to solve four K as a constant. So dividing both sides by X, we get that K is equal to our force divided by our X. So what this gives us is that K is equal to our maximum F divided by our maximum length of our stretched elastic core. Now that we have all this information, we are ready to go ahead and perform this integral and solve for our final velocity. But before we do that, let's go over our answer choices here. We have that our maximum final velocity for answer choice A is going to be 11.5 m per second or B it's 5. m per second. C is 133 m per second and D is 3. m per second. All right. Wonderful. So going back to it, let's go ahead and perform this integration right here while subbing in K to our elastic force equation as well. So we have work is equal to F max over X F our K, since it's just a constant, we can take it out. And then we are just going to integrate X with respect to X. So this will give us X squared over two And now we just have to evaluate it at our bounds, which is X I on the bottom and X F on the top. So what this gives us is F max divided by X F times X F squared over two minus X I squared over two. But as we can see at the beginning of our graph, there was no starting length for our stretch elastic chord. So this X I term is going to disappear and this gives us, gives us that work is also equal to F max times X F over two. Now, we have work is equal to this equation and work is equal to this equation. So let's set the two equal to one another and solve for our final velocity term. Let me just go and scroll down here just a little bit. All right. So we have that F max times X F over two is equal to M V F squared over two. Let me just fix my two here. Wonderful. What I'm gonna do is I'm gonna multiply both sides by two over M two over M. You'll see that on the left hand side here, the twos cancel out on the right hand side, the M over two and two over M cancel out. And then finally, I'm just gonna take the square root of both sides in order to get rid of this exponent here. So what this leaves us with, let me just scroll down just a little bit more here. What this leaves us with is that our final is equal to the square root of our maximum force times our maximum stretch of our last accord all divided by the mass of our object. So let's go ahead and plug in those values the square root of our maximum force. Well, looking at our graph here, our maximum force looks to be 20 newtons times our maximum stretch of our last accord, which we already know to be 0.2 m divided by our mass of 0.3 kg. And when you plug this into your calculator, we get a final velocity of 11. m per second, which corresponds to our final answer. Choice of a. Thank you all so much for watching. Hope this video helped see you all in the next one.

A 50 g rock is placed in a slingshot and the rubber band is stretched. The magnitude of the force of the rubber band on the rock is shown by the graph in FIGURE P9.56. The rubber band is stretched 30 cm and then released. What is the speed of the rock?

Verified Solution

Key Concepts

Hooke's Law

Work-Energy Principle

Kinetic Energy