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Solved: CALC In an experiment, one of the forces exerted
Chapter 7, Problem 30E(choose chapter or problem)
In an experiment, one of the forces exerted on a proton is \(\vec{F}=-\alpha x^{2} \hat{1}\), where \(\alpha=12 \mathrm{~N} / \mathrm{m}^{2}\). (a) How much work does \(\vec{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\vec{F}\) conservative? Explain. If \(\vec{F}\) is conservative, what is the potential-energy function for it? Let when
.
Equation Transcription:
Text Transcription:
vec{F}=-\alpha x^{2} \hat{1}
\alpha=12 \{~N} {m}^{2}
Vec F
Vec F
U=0
x=0
Questions & Answers
QUESTION:
In an experiment, one of the forces exerted on a proton is \(\vec{F}=-\alpha x^{2} \hat{1}\), where \(\alpha=12 \mathrm{~N} / \mathrm{m}^{2}\). (a) How much work does \(\vec{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\vec{F}\) conservative? Explain. If \(\vec{F}\) is conservative, what is the potential-energy function for it? Let when
.
Equation Transcription:
Text Transcription:
vec{F}=-\alpha x^{2} \hat{1}
\alpha=12 \{~N} {m}^{2}
Vec F
Vec F
U=0
x=0
ANSWER:
Solution 30E
Work done by a variable force, say F(x) is given by, …..(1), where
is the initial displacement and
is the final displacement of the body.
Given that,
Given,
So,
(a) When the body moves from the point (0.10 m,0) to (0.10 m, 0.40 m), there is displacement only along the y-axis, No displacement happens along the x-axis. The force acts along the x-axis. Hence force and displacement are perpendicular to each other.
This signifies that work done by the given force for the given displacement is zero, because both are perpendicular to each other.
(b) For the displacement (0.10 m,0) to (0.30 m,0), work done by the force is calculated as below.