If a 60 kg person on a 15 kg sled is pushed with a force of 300 N what will be a person's acceleration?

##4m/s^2[forward]##
Recall that Newton’s ##2^nd## Law is given by the formula:
##color(blue)(|bar(ul(color(white)(a/a)F_net=macolor(white)(a/a)|)))##
where:
##F_net=##sum of all forces acting on the object
##m=##mass
##a=##
For this problem we will set the positive directions to be east and down.
Start by breaking the variable ##m## into ##m_person+m_sled## since the acceleration of the person also depends on the mass of the sled.
##F_net=ma##
##F_net=(m_person+m_sled)a##
Assuming there is no friction involved the only acting force in the ##x## direction is the applied force. The forces acting in the ##y## direction are the normal and gravity forces. However when you add them together the force is equal to ##0N## so we will ignore them.
##a=F_app/(m_person+m_sled)##
Substitute your known values.
##a=(300N)/(60kg+15kg)##
##a=color(green)(|bar(ul(color(white)(a/a)4m/s^2color(white)(a/a)|)))##