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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    my answer key says 65 though

  3. anonymous
    • 5 years ago
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    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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