## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$F = 0.033~N$
We can find the time to reach maximum height. $t_{up} = \frac{v_y-v_{y0}}{a_y}$ $t_{up} = \frac{0 - (6.0~m/s)~sin(35^{\circ})}{-9.80~m/s^2}$ $t_{up} = 0.351~s$ The total time of the object's flight is double the time it takes to reach maximum height. $t = 2~t_{up} = 0.702~s$ We can find the horizontal acceleration. $x = v_{0x}~t+\frac{1}{2}a_x~t^2$ $a_x = \frac{2(x-v_{0x}~t)}{t^2}$ $a_x = \frac{(2)[(2.9~m)-(6.0~m/s)~cos(35^{\circ})(0.702~s)]}{(0.702~s)^2}$ $a_x = -2.23~m/s^2$ We can use the magnitude of the horizontal acceleration to find the magnitude of the electric force $F$. $F = m~a_x$ $F = (0.015~kg)(2.23~m/s^2)$ $F = 0.033~N$