anonymous 4 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

1. 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

2. anonymous

my answer key says 65 though

3. 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 $$v_f$$