Chapter 5 - Force and Motion-I - Problems - Page 119: 38b

$F_x = 28N$

$Fg = mg$ $Fg = 40kg \times 9.8m/s^2$ $a=1m/s^2$ Applying the 2nd law of motion: $F_{net} = Fgsin\theta-F_x$ $F_x = Fgsin\theta-F_{net}$ $F_x = Fgsin\theta-ma$ $F_x = (392\times sin(10^{\circ}))- (40\times 1)$ $F_x = 68 -40$ $F_x = 28N$