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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)|)))##

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