Is My Solution for Variable Force Particle Motion Correct?

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SUMMARY

The discussion centers on solving a physics problem involving a particle of mass m subjected to a variable force described by Fx = Fosin(cx). The correct position xmax where the force first reaches its maximum is determined to be xmax = π/2c. The particle's velocity at this position is derived using the work-kinetic energy theorem, yielding the formula v1 = sqrt(vo^2 + 2Fo/c*m). The calculations provided are confirmed as accurate, with a note on the importance of careful notation in mathematical expressions.

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Homework Statement


A particle of mass m starts from xo= 0m with vo> 0 m/s. The particle experiences the variable force Fx= Fosin(cx) as it moves to right along the x-axis, where Fo and c are constants.

a)At what position xmax does the force first reach a maximum value?

b) What is the particle's velocity as it reaches xmax? Give your answer in terms of m, vo, Fo, & c.

Homework Equations





The Attempt at a Solution



Here is my idea for part a:

Take the derivative and set it equal to 0
Fx'=Fo*c*cos(cx)
0=Fo*c*cos(cx)
cos(cx)=0
cx= pi/2
xmax=pi/2c Is this correct?

Then for part b:

I took the integral of Fosin(cx):

(Fo/c) * (-cos(pi/2) + 1) = Fo/c This is the work done.

Then using the work-kinetic energy theorem:

.5mvo^2 + Fo/c = .5 mv1^2 where v1 is the velocity at x max

.5vo^2 +Fo/c*m = .5v1^2
vo^2 + 2Fo/c*m = v1^2
sqrt (vo^2 + 2Fo/c*m) = v1
 
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It is a bit hard to follow your work, but everything seem to be correct.
 
A particle of mass m starts from x_0=0\,{\rm m} with v_0>0\,{\rm m/s}. The particle experiences the variable force F_x =F_0 \,{\rm{sin}}\left( {cx} \right) as it moves to the right along the x-axis, where F_0 and c are constants.


the vast majority of students trying to complete this problem on mastering physics will just copy and paste this... so I'll put it here...

the answer above is completely right (TIP, be careful with the parentheses)
 

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