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- Thread starter tasnim rahman
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Doc Al

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One way of seeing it is considering the energy a force insert to the system (or the energy per unit time):

P=[tex]\vec{f}[/tex]*[tex]\vec{}v[/tex] = 0

Another way is simply taking the derivative of the magnitude of the velocity (assume 2-D case):

d(v^2)\dt= d(v_x)^2\dt + d(v_y)^2\dt = 2(a_x*v_x + a_y*v_y) = 2[tex]\vec{a}[/tex]*[tex]\vec{v}[/tex]= 2\m([tex]\vec{f}[/tex]*[tex]\vec{v}[/tex]) = 0

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I believe that should be perfectly clear to everyone.alwaysperpendicular to the direction of motion does not change the magnitude of the velocity.

One way of seeing it is considering the energy a force insert to the system (or the energy per unit time):

P=[tex]\vec{f}[/tex]*[tex]\vec{}v[/tex] = 0

Another way is simply taking the derivative of the magnitude of the velocity (assume 2-D case):

d(v^2)\dt= d(v_x)^2\dt + d(v_y)^2\dt = 2(a_x*v_x + a_y*v_y) = 2[tex]\vec{a}[/tex]*[tex]\vec{v}[/tex]= 2\m([tex]\vec{f}[/tex]*[tex]\vec{v}[/tex]) = 0

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