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?

1 Answer
Apr 21, 2016

#4m/s^2["forward"]#

Explanation:

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=#acceleration

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)|)))#