Is My Solution for Variable Force Particle Motion Correct?

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The discussion revolves around solving a physics problem involving a particle under a variable force defined as Fx = Fosin(cx). For part a, the correct position xmax where the force reaches its maximum is derived as xmax = π/2c by taking the derivative of the force and setting it to zero. In part b, the work done by the force is calculated, leading to the velocity at xmax expressed as v1 = sqrt(vo^2 + 2Fo/c*m). Participants confirm the correctness of the calculations and emphasize the importance of careful notation. The solution is validated, indicating that the approach to solving the problem is sound.
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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)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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