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A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 5.10-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.If the box is initially at rest at x=0, what is its speed after it has traveled 14.0 m ?

Respuesta :

Answer:

[tex]v=7.62\ m.s^{-1}[/tex]

Explanation:

Given:

  • initial position of the box, [tex]x=0\ m[/tex]
  • final position of the box, [tex]x'=14\ m[/tex]
  • mass of the box under the force, [tex]m=5.1\ m[/tex]
  • initial speed of the box, [tex]u=0\ m.s^{-1}[/tex]
  • function of force, [tex]F(x)=18-0.53x\ [N][/tex]

where:

[tex]x=[/tex] distance in the +ve x-direction

We know:

[tex]F=m.a\\\Rightarrow a=\frac{F}{m}[/tex]

Now force change in force on the body:

[tex]F(x)=18-0.53(x'-x)[/tex]

[tex]F=18-0.53\times (14-0)[/tex]

[tex]F=10.58\ N[/tex]

Now the acceleration due to the force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{10.58}{5.1}[/tex]

[tex]a=2.0745\ m.s^{-2}[/tex]

Now using equation of motion:

[tex]v^2=u^2+2a.x'[/tex]

[tex]v^2=0^2+2\times 2.0745\times 14[/tex]

[tex]v=7.62\ m.s^{-1}[/tex]

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