CQ 1	If a car suddenly stops, the contents continue to move forward unless
	acted upon by an external force such as seatbelts or normal force of the
	dashboard.

CQ 6	Friction on wheels pushes car forward.  The engine pushes back
	on the ground, so by Newton's 3rd law, the ground pushes forward
	on the tire.

CQ 7	Each car is in motion, so by N1L, continues in straight-line motion
	until acted upon by an external force, the other car, which slows the
	car down.  Change velocity at finite acceleration takes time, Δt = Δv/a
	so even if the cars only have to go from 5m/s (~11mph) to 0m/s at 10g's, ~100m/s/s,
	Δt = (-5m/s) / (-100m/s/s) = .05s = 50ms
	Because cars are not truly rigid bodies, the front-end stops before the back-end.

CQ 9	The scale measures the spring force inside  the scale, which is the normal
	force the scale exerts up on the person.  This equals the weight of the person
	ONLY if the person is not accelerating up or down.

CQ 12	The acceleration is downward at 9.8 m/s/s the ENTIRE time the
	object is falling (up, top, or down).  If air resistance is negligible,
	the answer is still the same - b/c the force is that of gravitation and
	that is straight down always.

CQ 13	Even though the force of you on the wagon equals the force of the
	wagon on you, you can accelerate and the wagon can accelerate. What is
	required is a third body, the ground, to exert a net force, a push, on your
	feet forward (the action-reaction pair to you pushing backward on the ground).

CQ 12	g downward because the velocity is constantly changing, even it momentarily stopped

CQ 14	(a) scale reads the same as when stopped because a=0 so W=N
	(b) moving up with incr. speed implies upward a>0 implies N>W
	(c) in free fall, N=0!

CQ 16	Equal.  N3L

P 2(and what is its mass)	19.8N *(1 lb / 4.448 N) = 4.45 lb
	m = W/g = 19.8N / 9.8m/s/s = 2.02 kg

P 16	Net force in vert dir = Fair up on wings -mg
	= by N2L = ma = 0 b/c bird is hovering.  thus mg = .3N

P17	Forces are air pushin upward on bird and weight of bird (downward), so
	net upward force = Fair - mg
	but Fnet = 0 if bird is motionless, so Fair = mg = 0.30N
	[Since mg=0.3N, m = (0.3N) / (9.8m/s/s) = 0.031kg = 31 gm]

P22	2010kg elevator accelerates upward at 1.5m/s/s, thus the net force is ma=3015N.
	The cables exert an upward tension while the Earth pulls down with W=mg=19698N.
	Fnet = T-W = T - 19698, but N2L says Fnet = ma = 3015N, thus
	3015N = T-19700N, so T=22713N or 22700N or 22.7kN.

P 25	2kg stone (weight mg = 19.6N )
	lifted upward at constant upward acceleratio = 1.5m/s/s.
	Net force in vert dir = Fhand up on mass - mg =
	= by N2L = ma , so Fhand up on mass = mg + ma
	= m*(g+a) = (2kg)*(9.8+1.5m/s/s) = 22.6 N.

P 30	650N skydiver falls at constant speed (a=0). Parachute pulls
	up with 620N, thus there must be another 30N upward force, air drag.

P 31	FNET on HER = Ffloor up on HER_feet + 2*Farmrest up on HER_arms
		+Fseat up on HER_body - WEarth on HER
		= Ffloor up on HER_feet +2*25 + 500 - 600 = 0
		because a=0 and by N2L, Fnet=0, thus
		Ffloor up on HER_feet = 600-500-50=50N.

P 32	N3L action-rxn pairs: a=-b, (Earth down on bike) = -(bike up on Earth).
		N1L: hook up on bike = -(earth down on bike) - both on bike so Fnet=0

P61	(a) Force of hand on book plus opposing force of table on book (friction)
	plus vertical forces (normal=table up on bk and weight=earth down on bk)
	
	(b) After I stop pushing the book, the forces are exaclty the same as in (a)
	except for force of hand on book
	(c) After book stops sliding, only normal and weight act on book
	(d) The net force is non-zero in (a) provided you're speeding the book up
	and also in (b) because the friction is the only horizontal force, so the
	book is slowing down.
	(e) m=0.5kg => mg=4.9N and also N=4.9N b/c ay=0 so Fnety,y=0
	If friction coeff, mu,  is 0.4, then f=0.4*(4.9N)=1.96N, which is the
	only force acting in the horizontal direction, so the net force is 1.96N
	backwards.
	(f) If mu=0, then the only forces in part 9b) would be normal and weight,
	which cancel, so the net force would be zero, so the book would move at
	constant velocity (constant speed in straight line).

P64	The net force on the sleigh is zero (by N2L b/c the velo is constant)
	The physical forces add up to give Fnet of (T-friction) horizontally and
	(Normal-weight) vertically.  Both add to zero, so T=friction and N=m*g.
	(b) friction = mu*N, but T=friction = mu*N = mu*m*g, so T=mu*m*g OR
	mu = T/(m*g)

P83	Fnet,horizontal on both masses = T2	and N2L says = (m1+m2)*a
	Fnet,horizontal on m2 = T2 - T1	and N2L says = m2*a
	Fnet,horizontal on m1 = T1	and N2L says = m1*a
	So the first gives  T2 = (m1+m2)*a
	while the third,  T1 = m1*a.
	Dividing these two equations T1 / T2 = m1*a / ((m1+m2)*a) = m1 / (m1+m2)

P90	1400kg car (weight mg=13720N) is pulled with a rope that breaks at 2500N.
	If there is no friction, then the tension in the rope is the only horizontal
	force, so m*ax = 2500N at the breaking point.  Thus the maximum horizontal
	acceleration, ax = 2500N/1400kg = 1.786m/s/s.