A stretched spring stores 2.0 J of energy. How much energy will be stored if the spring is stretched three times as far?

Video transcript

Hi, everyone in this practice problem, we are given a hooks cord that is being stretched to a distance of D storing three joules of energy. Uh We're being asked to determine the energy stored when the court has a five D stretching or five times the stretch of the original distance D and the options given are a 25 Joles B Joles C 8.3 Joles and D 15 jos. So we want to consider this spring to be ideal for this particular practice problem and therefore, uh they will obey Hook's law. We denote the spring as S P or the subscript as P. And for a stretch spring, the potential energy US P U is denoting the potential energy and S P is denoting the spring US P is then going to be half multiplied by K multiplied by delta as squared and K is going to be the spring constant and delta S is going to be D A stretch. So let's take for the first instant which is going to be USB. Um For the first instance, delta S is going to equals to D and therefore USB will then be half K D squared just like so, and for the second, uh for the second case, when the court actually has five D stretching, then delta S will then be five D. And from there, our USB will then be half multiplied by K multiplied by delta S squared and delta S is five D. So USB is then going to be half K multiplied by five D squared. And that will give us USB equals to half K multiplied by 25 D squared just like. So, so the first case, we have the delta S equals to D and we have USB one to then be half K D squared. And for the second case, we have the delta S equals to five D for the five D stretching. And USB two will then be equals to half K 25 D squared just like. So, so the way we want to determine the energy stored when the court has a five D stretching is actually by taking the ratio of uh us P two and us P one. So we wanna take the ratio of us P two over us P one. And then that will actually give us half K multiplied by 25 multiplied by D squared, divided by half oops K multiplied by D squared. We can simplify the right side of our equation and pretty much cross out the half K and also cross out the D squared. And that will actually give us a ratio of us P two divided by us P one equals to 25. And rearranging this, we can know that us P two will equals to 25 multiplied by us P one or 25 times us P one. And that will give us 25 multiplied by us P one which is given, which is the energy stored in the stretch uh hooks cord which is three Jews. So 25 multiplied by three jules is going to be USB two, which is going to be 75 jules. So the energy stored for the second case, which is when the court has a five D stretching, the amount of energy stored in that is going to be 25 times the amount of the normal distance which is going to equals to 75 Joles. So 75 Joles is going to be corresponding to answer uh Option B in our answer choices and answer B with the energy stored of 75 Joel when the court has a five D stretching is going to be the answer to this practice problem. So if you guys have any sort of confusion on this particular practice problem, please still make sure to check out our other lesson videos and that will be all for this one. Thank you.