PHYSICS HELP PLEASE!! grade 11, thank you?

Question

on a level ice surface( coefficient of friction = 0.1) two boys stan facing each other 0.15m apart. one has a mass of 45kg and the other 65kg. the smaller boy pushes the bigger boy with a force of 55N for o.4s. At the en of the push how far apart will the boys be? THANK YOU

Answer 1

Fun problem

Every action has an equal and opposite reaction

m1 = mass of the small boy
m2 = mass of the big boy

Sum of the horizontal forces on the big boy = m2*a2 = F - friction
Sum of the horizontal forces on the small boy = m1*a1 = friction - F

The reason the signs are different for the small boy is because the force and friction are acting in the opposite direction.

Remember friction = u*m*g

a2 = F/m2 - u*m2*g
a1 = -F/m1 + u*m1*g

Use kinematics to figure out the distance traveled by each

I'm going to set up the coordinate system with the small boy at the origin and the big boy at +0.15m.

s1 = 1/2*a*t^2
s2 = s0 + 1/2*a*t^2

The acceleration will come out negative for the smaller boy and therefore he will have a negative displacement, vice versa for the bigger boy

Total distance apart = s2 - s1.

I hope I covered everything correctly. The assumption I made is that the Force exerted is stationary and not moving with the smaller boy. It is a difficult problem even for 11th grade.
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If the Force is moving with the smaller boy then the problem involves relative motion.
Our FBD of the big boy stays the same so acceleration remains the same;
a2 = F/m2 - u*m2*g

However, this acceleration is relative to the smaller boy

Just plug into a kinematics equation, s = s0 + 1/2at^2, and that would be the distance separating the two. Hope I could help.