Solution Found!
a. FIGURE shows the force Fx exerted on a particle that
Chapter 11, Problem 40P(choose chapter or problem)
a. FIGURE P11.40a shows the force \(F_{x}\) exerted on a particle that moves along the \(x\)-axis. Draw a graph of the particle's potential energy as a function of position \(x\). Let \(U\) be zero at \(x=0 \mathrm{~m}\).
b. FIGURE P11.40b shows the potential energy \(U\) of a particle that moves along the \(x\)-axis. Draw a graph of the force \(F_{x}\) as a function of position \(x\).
Equation Transcription:
Text Transcription:
F_x
x
x
U
x = 0 m
U
x
F_x
x
Questions & Answers
QUESTION:
a. FIGURE P11.40a shows the force \(F_{x}\) exerted on a particle that moves along the \(x\)-axis. Draw a graph of the particle's potential energy as a function of position \(x\). Let \(U\) be zero at \(x=0 \mathrm{~m}\).
b. FIGURE P11.40b shows the potential energy \(U\) of a particle that moves along the \(x\)-axis. Draw a graph of the force \(F_{x}\) as a function of position \(x\).
Equation Transcription:
Text Transcription:
F_x
x
x
U
x = 0 m
U
x
F_x
x
ANSWER:
Step 1 of 7
(a)
Using the given force position graph of a particle, we need to draw the potential energy versus the position graph of the particle if the initial potential energy at \(x=0\) is zero.
- To derive a relation between force and potential energy,
From the work-energy theorem, we know that work done during the process will be equal to the negative of the change in potential energy.
That is,
\(W=-\Delta U\)
Using work done as a product of force and displacement \(W=F \Delta x\)
\(F_{x} \Delta x=-\Delta U\)
On integrating,
\(U=-\int F_{x} \Delta x \ldots \ldots \ldots .1\)