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)|)))##