A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What are the block's b. Distance from equilibrium when the speed is 50 cm/s?

Video transcript

A little fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A student used a vertical spring mass system to investigate simple harmonic motion. A load of 250 g is attached to a vertical spring with a spring constant of 25. newtons per meter. The student observes the load has a repetitive movement back and forth through the equilibrium position at time T equals zero seconds. The load is 15 centimeters below the equilibrium position and moving straight up at a speed of 0.75 m per second. Calculate the loads position with respect to the equilibrium position when it reaches a speed of 0.25 m per second. So that's our end goal. So we're given some multiple choice answers. They're all in the same units of centimeters. Let's read them off to see what our final answer might be. A is plus or minus 14.4 B is plus or minus 16. C is plus 12.2 and D is negative 12.2. OK. So first off, let us note that the equilibrium position is placed at the origin point on the Y axis. The positive direction is upward. Let us also note that T equals zero seconds. So T equals zero seconds and the load is at position Y one equals negative 15. centimeters and has a speed. V one is equal to 0.75 m per second. Once the load reaches a speed of V two equals 0.25 m per second, it will be at position Y two. OK. So we also need to note that when the load reaches its maximum displacement position, the velocity will be zero because the mass is changing direction and at the equilibrium position, the velocity of the load will be at its maximum. Thus, we will consider the spring to be a load system, all of the forces will be conserved. Thus, the mechanical energy will also be conserved. So we can write that E one is equal to E two. Now let us consider that the equilibrium position as a reference level. If E one equals E two, we could write that K one plus you G one plus us, one is equal to K two plus UG two plus us two. So now we can plug in the values for K and U and note that K is the kinetic energy and U is the potential energy So one half MV, one squared plus mGy one plus one half K multiplied by delta Y zero plus Y one squared is equal to one half MV, two squared plus M gym multiplied by G multiplied by Y two plus one half K multiplied by delta Y zero plus Y two squared. OK. So now we need to rearrange this equation to set it equal to zero. OK. So let's get cracking at that. So one half MV, one squared plus M multiplied by G multiplied by Y one plus one, K multiplied by Y zero squared plus one half, multiplied by K multiplied by Y one squared plus K Y one multiplied by delta Y zero is equal to one half MV two squared plus M multiplied by G multiplied by Y two plus one half. Hey, multiplied by delta Y zero squared worse. So I'm gonna move it down below plus one half ky two squared plus ky two multiplied by delta Y zero. OK. Oh, that was a lot of writing. OK. We got this though. We got this. So let's continue simplifying shall we? And when we do, we should get that, it's one half multiplied by M multiplied by V one squared minus V two squared plus M multiplied by G multiplied by Y one minus Y two plus A multiplied by delta Y zero multiplied by Y one minus Y two plus one half multiplied by K multiplied by Y one squared minus one half multiplied by K multiplied by Y two squared is equal to zero. OK. So now we need to recall and apply Newton's second law at the equilibrium position. So the sum of FY is equal to MA Y is equal to zero. So we can write that negative K multiplied by delta Y minus MG is equal to zero. And then we can then go on to even say that negative K multiplied by delta Y zero is equal to MG. OK. And note that delta Y zero is negative. And let's call this equation two. And the equation one is our answer that we found when we set it equal to zero. OK. So now we combine equations one and two. So let's do that. So Y two squared is equal to Y one squared plus M divided by K multiplied by V one squared minus V two squared which isolating and solving for Y we get Y or solving for Y two, I should say. So Y two is equal to plus or minus the square root of Y one squared plus M divided by K multiplied by V one squared minus V two squared. So now we could substitute all of our known variables and to solve for Y two. OK. So Y two is equal to plus or minus the square root. So Y one is 0.15 meters squared plus our value for M. So the mass was zero point 2 0 kg divided by our spring constant, which was 25 newtons per meter. So that was given to us in the prom itself multiplied by B one which was 0.75 m per second squared minus V two, which was 0.25 meters per second squared. So when we plug that into a calculator, we should get that Y two is equal to plus or minus 0.166 m. But we need to convert this to centimeters. So then when we convert it to centimeters, it'd be plus or minus 16.6 centimeters. So the load reaches a speed of 0.25 m per second at the vertical position with respect to the equilibrium position at plus or minus 16.6 centimeters. So this is our final answer all ray. So let's go look at our multiple choice answers to see what the correct answer is. The correct answer is the letter B plus or minus 16. centimeters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.