Next, the particle is released from rest at x=0 . What are...

Next, the particle is released from rest at $x=0 .$ What are (l) its kinetic energy at $x=5.0 \mathrm{~m}$ and $(\mathrm{m})$ the maximum positive position $x_{\max }$ it reaches? (n) What does the particle do after it reaches $x_{\max } ?$ $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem $81,$ a $2.00 \mathrm{~kg}$ particle is released at $x=5.00 \mathrm{~m}$ with an initial velocity of $3.45 \mathrm{~m} / \mathrm{s}$ in the negative direction of the $x$ axis. (a) If the particle can reach $x=0 \mathrm{~m},$ what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive $x$ direction when it is released at $x=5.00 \mathrm{~m}$ at speed $3.45 \mathrm{~m} / \mathrm{s} .$ (b) If the particle can reach $x=13.0 \mathrm{~m},$ what is its speed there, and if it cannot, what is its turning point?

Next, the particle is released from rest at $x=0 .$ What are (l) its kinetic energy at $x=5.0 \mathrm{~m}$ and $(\mathrm{m})$ the maximum positive position $x_{\max }$ it reaches? (n) What does the particle do after it reaches $x_{\max } ?$
$$
\begin{array}{lc}
{\text { Range }} & \text { Force } \\
\hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\
2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\
3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\
8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\
11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\
12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0
\end{array}
$$
For the arrangement of forces in Problem $81,$ a $2.00 \mathrm{~kg}$ particle is released at $x=5.00 \mathrm{~m}$ with an initial velocity of $3.45 \mathrm{~m} / \mathrm{s}$ in the negative direction of the $x$ axis. (a) If the particle can reach $x=0 \mathrm{~m},$ what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive $x$ direction when it is released at $x=5.00 \mathrm{~m}$ at speed $3.45 \mathrm{~m} / \mathrm{s} .$ (b) If the particle can reach $x=13.0 \mathrm{~m},$ what is its speed there, and if it cannot, what is its turning point?

Key Concepts

Analysis of Motion in Piecewise Constant Force Fields

When forces act in different segments along an object's path, each with distinct magnitudes and directions, one must analyze each section separately. The work done over each interval is calculated using the constant force value, and these contributions are then summed to determine the overall effect on the object’s kinetic energy. This method allows for the piecewise determination of energy changes through intervals of variable force.

Potential Energy and Energy Conservation

In conservative force fields, potential energy represents the stored energy due to position, and it is defined so that its change equals the negative of the work done by the force. The conservation of mechanical energy (kinetic plus potential) allows one to analyze motion by setting initial energy equal to final energy, especially when calculating turning points where kinetic energy momentarily becomes zero.

Turning Point in Motion

The turning point is the position where the object's velocity becomes zero before it reverses direction. It corresponds to the location where all available kinetic energy has been converted into potential energy (in a frictionless or conservative system), making it a key indicator in energy-based motion problems.

Work-Energy Principle

This principle states that the net work done on an object by all forces equals the change in its kinetic energy. It is used to relate the work done by forces (calculated by integrating force over displacement) to changes in speed, which allows one to determine kinetic energy values or turning points in a motion problem.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated as one-half times its mass times the square of its velocity. This concept is central when using energy conservation techniques to determine speeds at particular positions or how energy transforms during motion.

Transcript

00:01 Now in the l sub part problem number 81 l sub part we have to calculate the kinetic energy at x equals to 5 meter so the potential energy at x equals to 0 meter was 11 joules now as the particle travels from 0 meters 3 meters what happens is that the particle gains kinetic energy and after reaching x equals to 5 meters since u is equal to 0 at x equals to 5 meters then its initial potential energy that is of 11 jules gets completely converted to kinetic energy.

00:44 So the total potential energy at x equals to 0 meters gets converted to kinetic energy at x equals to 5 meter.

01:13 Therefore, kinetic energy at 5 meter equals 11 joules.

01:27 Now the total kinetic energy at x equals to 5 meters, that is 11 joules, is not enough.

01:35 Why is it not enough? let us discuss.

01:38 In the else sub part, we need to calculate the x max.

01:42 You need to calculate the x max.

01:45 That the particle reaches.

01:46 Now the kinetic energy at x equals to 5 meters that is of 11.

01:49 Joules is not enough it is not enough to climb up and out of the well on the large x side that is x equal to x greater than 8 meters so but it allows it to reach a height of 11 at x equals to 10 .8 meters so x max will be equal to 10 .8 meters now why does x max come out to be 10 .8 meters it is because the kinetic energy at x equal to 5 meter equals 11 jules from x equal to 3 meters to x equal to 8 meters the force acting is 0 so there's no force acting between x equals to 3 meters to x equals to 8 meters and from x equals to 8 meter to x equal to 11 meters the force acting is f3 okay the f3 force is and its magnitude is minus 4 newton i cap so it is acting in the opposite direction of motion now as it is acting in the opposite direction it is doing negative work and as it is doing negative work so it will change the kinetic energy to potential energy now what will happen is that the total kinetic energy suppose from x equals to 3 meters to x equals to 8 meters since there is no force the kinetic energy will not be converted to potential energy at that point it will only get converted from x equals to 8 meters to x equals to 11 meters so the kinetic energy is 11 joules which will be converted by the work done by the force f3 so that will be equal to 11 equals minus 4 into d or delta x okay we can also like this that as delta x so delta x now delta x now delta x will be equal to minus minus 11 by 4 that comes out to be 2 .8 meter now this delta x is from 8 meters so the actual value actual value of x max will be 8 meter plus 2 .8 meter that is equal to 10 .8 meter that we discussed about it earlier now the nth sub part asks us that what does the particle do after it reaches x max now the nth sub part is what does the particle do after it reaches x max so what happens is that it falls back down and rises back up the small x slope until it comes back to its original position okay so at x equals to 10 .8 meters which is x max this is x max now at that at this point it momentarily stops and then it is accelerated to the left...

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