Find Fmax for a 450g Particle Moving Along the x-Axis | Ex11.15

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To find Fmax for a 450 g particle moving along the x-axis, the maximum force occurs at 1 meter on the force versus distance graph. The particle's velocity increases from 4.0 m/s to 7.5 m/s between 0 m and 2 m. The work-energy theorem is applicable, where the change in kinetic energy (ΔKE) equals the net work done (WNet). By calculating ΔKE using the initial and final velocities, and considering the area under the force curve, the correct maximum force is determined to be approximately 9.05625 N. This approach effectively utilizes the relationship between force, mass, and acceleration to solve the problem.
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A 450 g particle moving along the x-axis experiences the force shown in Figure Ex11.15. The particle goes from vx = 4.0 m/s at x = 0 m to vx = 7.5 m/s at x = 2 m. What is Fmax?

Where the maximum peak on the force by distance graph is at 1m

F=ma
v=d/t
a=v/t


I have no idea how to answer this. I tried to find the acceleration then multiply it by the mass

i got 7.2 n which was wrong
 
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Use the work energy theorem
ΔKE = WNet
You can calculate the change in kinetic energy using the given numbers. Since only F is acting on the particle, F is the net force. The work done by F is the area under the curve, so ...
 
.5mv^2-.5mv^2 using 7.5 and 4 as the velocities. thank you very much

9.05625 btw
 
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