Two boys stand on the surface of a frozen lake where μs = 0.15 and μk = 0.10. The 30.0 kg boy shoves the 40.0 kg boy with a horizontal force of 50.0 N, 0.0°. Find the resulting acceleration of each boy.

Answer 1

Let A=30 kg boy
B=40 kg boy
Force = 50 N

For boy A:
Static Friction = mgμs=30*9.8*0.15=44.1N
< 50N =>
Boy A is set in motion.
Net force after kinetic friction
= 50N-30*9.8*0.1N=5.9N
Acceleration = F/m = 5.9/30=0.69 m/s² (backwards)

For B, the 40kg boy
Static friction = mgμs=40*9.8*0.15
=58.8 > 50N, so boy remains stationary.

That should teach the small boy a lesson!

Answer 2

The above response is good, except this part:

net force after kinetic friction=50N-30*9.8*.1=20.6N

Acceleration = F/n = 20.6/30 = .686666666 m/s^2

The rest of MathMate's solution looks good, but hope this helped!