## Engineering Mechanics: Statics & Dynamics (14th Edition)

$162N\cdot s$
The required impulse can be determined as follows: First, we apply the principle of impulse and momentum in the y-direction $mv_{y_1}+\Sigma \int ^{t_2}_{t_1} F_y dt=mv_{y_2}$ We plug in the known values to obtain: $0+Nt-6(9.81)t=0$ $\implies N=58.86N$ Now, we apply the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma \int ^{t_2}_{t_1}F_x dt=mv_{x_2}$ We plug in the known values to obtain: $0+\int ^{t_2} _{t_1} F_x dt-(0.5)(9.81)(75)(0.4)=75(0.2)$ $\implies \int F_x dt=162N\cdot s$