Biologia

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Cap 6
1. We do not consider the possibility that the bureau might tip, and treat this as a purely
horizontal motion problem (with the person’s push
F
in the +x direction). Applying
Newton’s second law to the x and y axes, we obtain
, max
0
s
N
F f ma
F mg
−
−
respectively. The second equation yields the normal force FN = mg, whereupon the
maximum static friction is found to be(from Eq. 6-1) fs,max s mg . Thus, the first
equation becomes
F mg ma s −0
where we have set a = 0 to be consistent with the fact that the static friction is still (just
barely) able to prevent the bureau from moving.
(a) With 
s 0.45 and m = 45 kg, the equation above leads to F = 198 N. To bring the
bureau into a state of motion, the person should push with any force greaterthan this
value. Rounding to two significant figures, we can therefore say the minimum required
push is F = 2.0 102 N.
(b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly
F 1.2102 N.
2. To maintain the stone’s motion, a horizontal force (in the +x direction) is needed that
cancels the retarding effect due to kinetic friction. Applying Newton’s second to the xand y axes, we obtain
0
k
N
F f ma
F mg
−
−
respectively. The second equation yields the normal force FN = mg, so that (using Eq. 6-2)
the kinetic friction becomes fk = 
k mg. Thus, the first equation becomes
F −kmg ma 0
where we have set a = 0 to be consistent with the idea that the horizontal velocity of the
stone should remain constant. With m = 20 kg and 
k = 0.80, wefind F = 1.6 102 N.
3. We denote
F
as the horizontal force of the person exerted on the crate (in the +x
direction),

fk is the force of kinetic friction (in the –x direction), N F is the vertical
normal force exerted by the floor (in the +y direction), and mg is the force of gravity.
The magnitude of the force of friction is given by fk = 
kFN (Eq. 6-2). Applying Newton’s
secondlaw to the x and y axes, we obtain
0
k
N
F f ma
F mg
−
−
respectively.
(a) The second equation yields the normal force FN = mg, so that the friction is
0.3555 kg(9.8 m/s2 ) 1.9 102 N . k k f mg 
(b) The first equation becomes
F mg ma k −
which (with F = 220 N) we solve to find
a F
m
g k −0.56 m/ s2 .
4. The free-body diagram for the player is shown next.N F

is the
normal force of the ground on the player, mg is the force of gravity,
and

f is the force of friction. The force of friction is related to the
normal force by f = 
kFN. We use Newton’s second law applied to
the vertical axis to find the normal force. The vertical component of
the acceleration is zero, so we obtain FN – mg = 0; thus, FN = mg.
Consequently,
2 
470 N0.61.
79 kg 9.8 m/s k
N
f
F

5. The greatest deceleration (of magnitude a) is provided by the maximum friction force
(Eq. 6-1, with FN = mg in this case). Using Newton’s second law, we find
a = fs,max /m = 
sg.
Eq. 2-16 then gives the shortest distance to stop: |x| = v2/2a = 36 m. In this calculation,
it is important to first convert v to 13 m/s.
The +x direction is“downhill.’’
Application of Newton’s second law to the x- and y-axes leads to
sin
cos 0.
k
N
mg f ma
F mg


−
−
Solving these along with Eq. 6-2 (fk = 
kFN) produces the following result for the pig’s
downhill acceleration:
sin cos . k a g −
To compute the time to slide from rest through a downhill distance , we use Eq. 2-15:
v t at t 
0 a
1 2
2
2 .
We denote thefrictionless ( 
k = 0) case with a prime and set up a ratio:
t
t
a
a
a
′a


2 ′
2


/
/
which leads us to conclude that if t/t' = 2 then a' = 4a. Putting in what we found out
above about the accelerations, we have
sin 4 sin cos . k g g −
Using = 35°, we obtain 
k = 0.53.
6. We first analyze the forces on the pig of mass m. The incline angle is ....
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