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shrutij

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## Homework Statement

A 2.40 kg particle moving along the x-axis experiences the force shown, where Fmax=6.41 N, Fmin=-6.41 N, d1=1.77 m, d2=3.54 m, d3=5.31 m, and d4=7.08 m. At x=0.00 m the particle's velocity is 4.54 m/s. What is its velocity at x=3.54 m?

http://capa.physics.mcmaster.ca/figures/kn/Graph11/kn-pic1114.png

## Homework Equations

Work energy theorem

## The Attempt at a Solution

W=KE

Since work is just the area under the force position graph, 1/2*b*h=1/2*m(vf^2-vi^2).

I got the area of the triangle= 3.54*6.41*0.5=11.35 J. when this was equated to KE, and isolated for vf^2= 2W/m +vi^2.

from this, i got vf=5.48 m/s, which is wrong.

Any ideas?

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