A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 6.20 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion with a varying force.

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Question:

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 6.20 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion with a varying force.

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QUESTION: If the box is initially at rest at x=0, what is its speed after it has traveled 12.0 m ?Answer:The velocity is 7.57/s.Explanation:Units aside, [tex]F(x)[/tex] can be written as [tex]F(x) = 18-0.530x[/tex].The work done by this force must equal the kinetic energy of the box: [tex]$\int_0^{12} F(x)dx = \frac{1}{2}mv^2 $[/tex][tex]$\int_0^{12} 18-0.530x\: dx = \frac{1}{2}(6.2)v^2 $[/tex][tex]$[18(12)-0.530\frac{(12)^2}{2}]- [18(0)-0.530\frac{(0)^2}{2}] = \frac{1}{2}(6.2)v^2 $[/tex][tex]$177.84 = \frac{1}{2}(6.2)v^2 $[/tex][tex]\boxed{v= 7.57m/s}[/tex]which is the velocity of the box after it has traveled 12 meters.

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