A 100 g particle experiences the one-dimensional, conservative fo...

Question

A 100 g particle experiences the one-dimensional, conservative force Fx shown in FIGURE P10.60.b. Suppose the particle is shot to the right from x=1.0 m with a speed of 25 m/s. Where is its turning point?

Accepted Answer

Hi, everyone in this practice problem, we are being asked to determine the position of a balls turning point. We will have a small ball with a mass of 50 g experiencing a conservative force of X in one dimension. If the ball is initially at X equals to 2.0 m and given a velocity of 15 m per second to the left, we are being asked to determine the position of the ball's turning point by finding the point where the kinetic energy of the ball is zero and the potential energy is at a maximum. We are given a diagram or a figure or a graph of X versus Y both in meters which depict the position of the small ball. The options given for the position of the ball starting points are a 3.29 m B 6.64 m C 2.56 m and D 4.58 m. So when the ball is at its turning point, its velocity, it's going to be zero. So therefore, the kinetic energy also will become zero. Therefore, the law of conservation of energy which we're going to employ in order for us to solve this problem will then be equal to the initial pe for potential energy plus the initial ke or kinetic energy will be equal to the final pe plus the final ke. So this is the law of the conservation of energy. And I'm going to just represent the initial pe with U I for the initial potential energy, initial Kek I for the initial kinetic energy and UF plus KF for the final potential and kinetic energy, as we have discussed previously, the kinetic energy at its turning point is going to be equal to zero. So KF will be equals to zero. So this conservation of energy will just becomes U I plus K I will be equal to UF just like. So rearranging this, we will get K I to then be equal to UF minus U I or essentially equals to delta U. So delta U will then be equal to the initial kinetic energy which is going to be half multiplied by M multiplied by V I squared. We can actually substitute the values that we are given in the problem statement because we are given the mass of the small ball and the velocity. So let's substitute that into this equation that we have so that we can determine the difference in potential energy. So then we will have half multiplied by M is going to be 50 g and changing that or converting that 2 kg. This will be 0.05 kg just like. So multiply that with uh V I which is 15 m per second squared just like. So and in calculating all of this, we will obtain the delta U value to be equal to 5.63 jus just like so, so we're going to hold on to that for now. And now we want to solve an expression or solve for an expression for the change in potential energy of the ball from X equals 2. m that traveled at a distance of X. So we want to utilize the equation of delta U to be equal to negative F multiplied by X which is the force multiplied by the distance traveled. And that will essentially be obtained from the graph given in the problem statement. So in the graph at X equals to 2 m, the uh Y axis is giving negative 10 Newton or negative 10. So in this case, the F is going to be negative 10. So therefore, we are going to employ that. So we have negative multiplied by F which is negative 10 Newton multiplied by X which is going to be what we are interested to find. I'm just gonna change it to cursive so that it's easier for us to see and not confuse us with a multiplication symbol. And then therefore, we will have a formula for X to then be equal to delta U divided by, than Newton which in this case, the delta U is obtained previously to be 5.63 Joles divided by than Newton. And that will give us the AX which is going to be equal to 0.56 m. Hence, we can then determine the final position of the ball or XF will then be equal to the initial, which is 2 m plus the displacement X here that we have, which is going to be 0.56 m and that will be obtain a total or the final position of the ball to then be equal to 2.56 m which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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A 100 g particle experiences the one-dimensional, conservative force Fx shown in FIGURE P10.60.b. Suppose the particle is shot to the right from x=1.0 m with a speed of 25 m/s. Where is its turning point?

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