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Take Two: problem using Work-Energy Theorem
The only force acting on a 1.5 kg body as it moves along the positive x-axis has an x component Fx = - 6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s.
(a) What is the velocity of the body at x = 4.0 m?
(b) At what positive value of x will the body have a velocity of 5.0 m/s?
WebAssign says that this problem uses concepts from the sections on Work Done by a Spring Force and Work and Kinetic Energy. I did not see how spring concepts were relevent, so I ignored them, though when I evaluated the integral, it looked a lot like a spring.
I got part (b) correct, but I'm down to my last response for part (a). Can someone tell me if this looks all right?
[tex]W=\int{F(x)dx} = \int{-6xdx} = -6\frac{x^2}{2} = -3x^2[/tex]
I evaluated the integral between x_i=3 and x_f=4 to get [tex]W = -3(4^2)-(-3)(3^2) = -21[/tex]
[tex]W=\Delta K=\frac{m}{2}(v_f^2-v_i^2)[/tex]
so [tex]\frac{2W}{m}+v_i^2=v_f^2[/tex]
so [tex]v_f = \sqrt{36} = 6m/s[/tex]
Thanks!
The only force acting on a 1.5 kg body as it moves along the positive x-axis has an x component Fx = - 6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s.
(a) What is the velocity of the body at x = 4.0 m?
(b) At what positive value of x will the body have a velocity of 5.0 m/s?
WebAssign says that this problem uses concepts from the sections on Work Done by a Spring Force and Work and Kinetic Energy. I did not see how spring concepts were relevent, so I ignored them, though when I evaluated the integral, it looked a lot like a spring.
I got part (b) correct, but I'm down to my last response for part (a). Can someone tell me if this looks all right?
[tex]W=\int{F(x)dx} = \int{-6xdx} = -6\frac{x^2}{2} = -3x^2[/tex]
I evaluated the integral between x_i=3 and x_f=4 to get [tex]W = -3(4^2)-(-3)(3^2) = -21[/tex]
[tex]W=\Delta K=\frac{m}{2}(v_f^2-v_i^2)[/tex]
so [tex]\frac{2W}{m}+v_i^2=v_f^2[/tex]
so [tex]v_f = \sqrt{36} = 6m/s[/tex]
Thanks!
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