FIGURE EX9.20 is the force-versus-position graph for a particle m...

Question

FIGURE EX9.20 is the force-versus-position graph for a particle moving along the x-axis. Determine the work done on the particle during each of the three intervals 0–1 m, 1–2 m, and 2–3 m.

Accepted Answer

Hi everyone in this practice problem, we're being asked to calculate the work done on an object where we will have the force applied to that object moving in a horizontal direction, varying with its position. The force is shown in the figure as a graph of FX versus X where FX is in Newton and X as in centimeter where from 0.0 to 2.5 centimeter FX is constant at two Newton at 20.2 point five centimeter, 2.10 centimeter. The FX is increasing from two Newton to five Newton and from 50.10 centimeter to 15 centimeter. D FX is decreasing from five Newton to zero Newton. We're being asked to calculate the work done on the object first between 0 to 2.5 centimeter second between 2.5 and 10 centimeter and third between 10 and 15 centimeters. The options given are a first zero, Jule, 2nd 0.3 Joles and third negative 0.3 Joles B 1st 0. Joles 2nd 0.26 Joles and 3rd 0. Jules C first zero, Joles second three Joles and third negative three Joles and D first five Joles, 2nd 26 Joles and 3rd 15 Joles. So the object is going to be moving along the horizontal axis or x axis by a force that varies with position. The work done by FX on the object as it moves from X I to X F can be expressed as a W which is going to be work equals to the in the girl from X I to X F of FX multiplied by D X just like. So the integral from X I to X F of FX D X is going to actually be equal to the area under the force position curve between X I and X F. So in this case, let's start from the first uh part which is the work done on the object between zero and 2.5 centimeter. So for between X I equals zero centimeters, N X F equals 2.5 centimeters, the work done is actually going to be represented by this rectangle right here. So in that case, uh W will then equals the integral from zero, which is I am going to convert everything into meter just so that it's easier for us to actually calculate. So the integral from zero m to 2.5 centimeter which is going to be 0.0 25 m of FX T X is then going to equals to the area of the rectangle right here. So that will be the width of the rectangle which is going to be 2.5 centimeter or in this case, then that will actually be equals to 2.5 times 10 to the power of minus two m multiply that by the length which is going to be to Newton and that will give us the work of five times 10 to the power of negative two just like. So, so the work done between X I equals zero centimeter and X F equals 2. centimeter is five times 10 to the power of negative two jules which will equals to 0.5 jules. Awesome. So now let's move on to the second part which is the work done between X I equals 2.5 centimeter and X F equals 10 centimeters. So using the same formula W will equals due to integral from 0.0 25 m to 0.1 m of FX multiplied by D X del essentially equals to first the area of this rectangle right here and add that with the area of our triangle on top of the rectangle right here. So let's start by calculating the area of our rectangle. So the width of the rectangle is going to be from 2.5 to centimeter, which means that it will be 7.5 centimeters or 7. times 10 to the power of negative two m. And then we want to multiply that by the length which is going to be two Newton. And then we wanna add that with our triangle, which in this case, then it will be half multiplied by the base, which is the same as the width of the rectangle, which is 7.5 times 10 to the power of negative two m. And we wanna multiply that by the height of our triangle which is from five Newton to two Newton, which is essentially three Newton. But I'm just gonna write down five minus two Newton just like. So, so then that will actually give us an A W or a work done value total of 2.63 times 10 to the power of negative one Joel or in this case, that will equals to 0.26. So that will be the second part of the problem. And now on to the last part or the third part which is the work done between 10 centimeter to 15 centimeters. So that mean X I will equals to 10 centimeter and X F will equals to 15 centimeter. And that will be then uh the W of the integral or the W will then equal to the integral from 0.1 m to 0.15 m of FX D X. And that integral will equals to the area of this triangle that we have right here just like. So, all right, so the triangle will have be CAL with then can be calculated by a multiplying half with the base of the triangle which is from 10 centimeters to 15 centimeter, which means that it will be five centimeter or five times 10 to the power of negative two m. And we want to multiply that by the height of our triangle which will be for from five Newton to zero, Newton. So I'm gonna write down five Newton and that will give us the work done value of 1.25 times 10 to the power of negative one or essentially equals to 0. Jews or rounding it up 0.13 Jules. So that will essentially be all the answer to this practice problem. So first between zero centimeter and 2.5 centimeter, we have the work done to be 0.5 jules and between 2.5 centimeters and centimeters, we have the work done to be 0.26 jules and lastly between 10 centimeter and 15 centimeter, we have the work done to be 0.13 jules which will actually correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that will essentially be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

FIGURE EX9.20 is the force-versus-position graph for a particle moving along the x-axis. Determine the work done on the particle during each of the three intervals 0–1 m, 1–2 m, and 2–3 m.

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