習題二
8.36. Let the initial motion of the cart be in the  x direction. The x component of the velocity will
at 2
atf3
dt  vx,i 
. So,
3m
0 m
tf
change according to
tf
vx,f  vx,i   ax (t )dt  vx,i  
ti
vx ,f (t  3.5 s)  (0.23 m/s) 
the
time
vx ,f  t   vx ,i 
vx ,f (t )  vx ,i 
at 3
3m
and
(0.0200 N/s2 )(3.5)3
 0.11 m/s. We can use the same expression to find
3(2.34 kg)
at
which
1/3
 3mvx ,i 
at 3
0t 

3m
 a 
the
cart
stops:
1/3
 3(2.34 kg)(0.23 m/s) 


(0.0200 N/s 2 )


 4.3 s.
10.56. The vertical height of the ground rises as the horizontal position moves to the right, according to
yslope ( x)  x tan(slope )
(1)
1
1
The vertical position of the ball is given by y(t )  yi  vy ,i t  a y t 2  yi  vi sin( )t  g t 2 . We wish
2
2
to relate this to the x direction. Ignoring air resistance the x component of the velocity is constant, so
t 
x
x

.
vx,i vi cos( )
Inserting this into the equation we obtained from the y direction, we have

 1 

x
x
y ( x)  yi  vi sin( ) 
  g

 vi cos( )  2  vi cos( ) 
2
or
y ( x)  yi  x tan( ) 

1 
x
g

2  vi cos( ) 
2
The ball will strike the ground when y ( x)  yslope ( x ). Setting the expressions in equations (1) and (2) equal
to each other yields
g
x 2  x(tan( )  tan(slope ))  yi  0
2v cos 2 ( )
2
i
Which can be solved for x using the quadratic formula, to obtain
x
2
cos2 ( )  2
8 gvi2
 2vi2 (tan(slope )  tan( ))    14.6 m. This is the
2vi (tan(slope )  tan( )) 
2
2 g 
cos ( )

horizontal distance that the ball travels, but we are asked for the distance d along the
incline: d 
x
(14.6 m)

 16 m.
cos(slope ) cos(20 )
(2)
10.87. (a)
F
x
t 
F
k
Ecx
Call
your
initial
direction
of
motion
the
x
direction.
Then
 k mg  max  ax  k g. Since this is a constant, we can use the kinematic equation
(vc x ,f  vc x ,i )
ax

 (0)  (27 m/s) 
(0.80)(9.8 m/s 2 )
 3.4 s.
(b) Following the same procedure as in part (a) we have t 
(c) We use vc2 x,f  vc2 x,i  2ax x and find x 
vc2 x ,i
2a x

(vc x,f  vc x,i )
vc2 x ,i
2k g
ax


((0)  (27 m/s))
 11 s.
(0.25)(9.8 m/s 2 )
(27 m/s) 2
 1.5  102 m.
2(0.25)(9.8 m/s 2 )
10.90. (a) Take the direction of motion to be the positive x-direction.
𝑣𝑥 = 𝑐𝑜𝑛𝑡𝑎𝑛𝑡 → 𝑎𝑥 = 0 → ∑ 𝐹𝑥 = 0
𝑓𝑘 = 𝜇𝑘 𝑁 = 𝜇𝑘 [𝑚𝑔 − 𝑇sin30]

𝑇cos30 = 𝑓𝑘

𝑇=

𝑇=
𝜇𝑘 mg
cos30+𝜇𝑘 sin30
0.2×20×9.8
0.866+0.2×0.5
= 40.6 N.
⃗⃗⃗⃗ = (𝑇𝑐𝑜𝑠30 ) ∙ 15 = 527.1 𝐽.
(b) 𝑊𝑥 = ⃗⃗⃗
𝑓𝑘 ∙ 𝛥𝑥
10.92. Choose the x axis to be horizontal such that the push we exert has a component that is in the  x
direction and let  y point vertically upwards. We write the sum of all x and y components of forces
separately:
F
F
x
s
 Fpbc x  Fsbx
 max
y
 Fsbn  Fpbc y  FEbG y  ma y  0
Or equivalently, assuming we are very close to getting the book to slip such that Fsbs   s Fsbn ,
Fpbc sin( )  s Fsbn  max
(1)
F  F cos( )  mg  0
(2)
n
sb
c
pb
Solving equation (2) for the normal force and inserting this into equation (1), we obtain
ax 
Fpbc
m
(sin( )   s cos( ))   s g
A negative acceleration in the above equation is, of course, nonsense because from the beginning we defined
the  x axis such that the push had some component along it. A negative acceleration in the equation above
means we are breaking one of the assumptions that we made, namely that we are maxing out the force of
static friction. Clearly, if   0 (for example) the surface could not possibly be exerting the maximum
possible force of static friction. Positive accelerations or an acceleration of zero make perfect sense. So
clearly the term (sin( )  s cos( ))  0, with equality only holding if the coefficient of static friction is
zero. The problem says we can push with an arbitrarily large force, and the book will not move unless the
angle is larger than some minimum value. If we take the limit as our force becomes extremely large, the
(3)
quantity (sin( )  s cos( )) can be arbitrarily small and the acceleration will still be positive. So in this
limiting case we set
(sin( )   s cos( ))  0    tan 1 (  s ). Thus   tan 1 (  s ) or no magnitude of
pushing force will make the book move.
10.105. Choose the  x axis to point down the incline, and the  y axis to be perpendicular to the
incline, with an upward component. We refer to the top layer of snow with a subscript “t”, and the surface of
snow beneath it with a subscript “s”. The sum of the x and y components of all forces yields
F  F
F  F
G
Etx
x
y
n
st
 Fstsx  max  0  mg sin( )  s Fstn
(1)
 may  0  F  mg cos( )
(2)
F
G
Eby
n
st
In equation (1), in setting the force of static friction equal to  s Fstn , we have assumed that the
force of friction is as large as it can get before the top layer slips. So this condition does
correspond to the largest possible angle before an avalanche occurs. Combining equations (1)
and (2) yields sin( )  s cos( ) or
  tan 1 (  )  tan 1 (3.7)  75.