anonymous
  • anonymous
A 2.0-kilogram body is initially traveling at a velocity of 40. meters per second east. If a constant force of 10. newtons due east is applied to the body for 5.0 seconds, the final speed of the body is
Physics
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anonymous
  • anonymous
A 2.0-kilogram body is initially traveling at a velocity of 40. meters per second east. If a constant force of 10. newtons due east is applied to the body for 5.0 seconds, the final speed of the body is
Physics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
u = 40m/s F = ma a= F/m a = 10/2 = 5 m/s^2 v = u + at v= 10 + 5*5 = 35 m/s
anonymous
  • anonymous
my answer key says 65 though
anonymous
  • anonymous
Why don't we use the principles of Impulse and momentum? Force is defined as\[{\bf \vec F} = {d \vec p \over dt}\]where momentum \(p\) is defined as\[\vec p = m*\vec v\]Now let's define impulse as\[{\bf I} = \int\limits\limits_{t_1}^{t_2} {\bf \vec F} dt = \int\limits\limits_{t_1}^{t_2} {d \vec p \over dt} dt = \int\limits_{p_1}^{p_2} d \vec p = \Delta \vec p\]We know the force applied, the initial velocity, and the time the force is applied for. Let's manipulate our expression for impulse to make use of these three things. \[\vec {\bf F}~t = \Delta p\](Note \({\bf \vec F} ~ t\) is after integration of the second term in the above equation.)Expanding further yields, \[{\bf \vec F}~t = m_f* \vec v_f - m_i * \vec v_i\]Since mass doesn't change\[{\bf \vec F} ~ t = m(\vec v_f - \vec v_i)\]This can be manipulated to solve for \(v_f\)

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