Solutions to Mid-Term Exam for GP I,
SNME, Fall 2011.
1.
5% + 10% + 10%
The figure shows a force lifting two blocks connected by a heavy
uniform rope.
(a) Find the acceleration of the system.
(b) Find the tension at the top of the rope.
(c) Find the tension at the mid-point of the rope.
Solution
(a)
F  mtot g  mtot a

a
F
200 N
g 
  9.80 m / s 2   3.53 m / s 2
mtot
 6.00  4.00  5.00 kg
(b)
6.00 kg block:

F  m6 g  Tt  m6a
Tt  F  m6  g  a 
 200 N   6.00 kg    9.80 m/s 2  3.53 m/s 2   120 N
(c)
5.00 kg block:

Tm 
Tb  m5 g  m5a
Tb  m5  g  a    5.00 kg    9.80 m/s2  3.53 m/s2   66.7 N
1
Tt  Tb   93.3 N
2
2. 10% + 10%
A block is attached to a rod by two strings. When
the system rotates about the rod as shown in the
figure, the tension in the upper string is 80.0 N.
(a) What is the tension in the lower string?
(b) Find the number of revolutions per minute at
which the lower string goes slack.
Solution
The block moves in a horizontal circle of radius r  (1.25 m) 2  (1.00 m) 2  0.75 m ,
makes an angle  with the vertical, cos  
1.00 m
1.25 m
  36.9° .
so that
(a)
y:
Tu cos  Tl cos  mg  0
Tl  Tu 
mg
cos 
 80.0 N 
(4.00 kg)(9.80 m/s2 )
cos36.9°
 31.0 N
(b)
x:
Number of revolutions per unit time:
Tu 
gives
n
v2
r
r Tu  Tl  sin 
(0.75 m)(80.0 N  31.0 N)sin 36.9°

 3.53 m/s
m
4.00 kg
v
Tl  0
Tu  Tl  sin   m
1
2
n
v
2 r

1
2
Tu  Tl  sin 
rm
mg
cos 
Tu sin 
1

rm
2
1
g sin 

r cos  2
 0.498 rev / sec  29.9 rev / min
 9.80 m / s  tan 36.9
2
 0.75 m 
3.
15%
A skier starts at the top of a very large, frictionless
snow ball, with a very small initial speed, and skis
straight down the side (see figure).
At what point does she lose contact with the snow
ball and fly off at a tangent?
Solution
Losing contact means n  0 .
y:
mg cos   n 
n0
m v22
R
v22  Rg cos 

Energy conservation:
1
mgR  mgR cos   m v22
2

1
mgR  mgR cos   mgR cos 
2
cos  
2
3
  48.2
4. 20%
A rough plank with an angle  to the horizontal has a varying coefficient of friction
given by s  k  A x , where A is a positive constant and x is the distance along
the plank measured from its bottom. A work shoves a box up the plank so that it it
leaves the bottom at speed v0.
Show that when the box first comes to rest, it will remain at rest if
v02 
3g sin 2 
A cos 
Solution
Let the box comes first to rest at x  X . It will stay at rest only if
s mg cos   mg sin 
i.e.,
X
Initial energy of box:
E0 
1 2
mv0
2
Energy of box when it comes to rest:
E1  mgX sin 
Work done by friction:
X
1
W   Amg cos   x dx   A X 2mg cos 
2
0
Energy-work theorem:
1
1
mgX sin   mv02   A X 2mg cos 
2
2
v02  A X 2 g cos   2 gX sin 
For the box to stay at rest, we must have
sin 
 sin  
v  A
sin 
 g cos   2 g
A cos 
 A cos  
2
2
0
sin 2 
v  3g
A cos 
2
0
sin 
A cos 
5. 20%
Due to air drag from Earth’s atmosphere, the radius of a satellite decreases from r
to r  r, where 0  r
r.
(a) Show that the increase of the orbital speed is v  
r
2
G mE
, where mE
r3
is the mass of the Earth.
(b) Show that the work done by the force of the air drag is W  
G mE m
r ,
2r 2
where m is the mass of the satellite.
Solution
(a) For a circular orbit around Earth,
v2
m
 G 2E
r
r


v  v 

v 
r
2
v
G
mE

G
r  r
G
mE
r
m
G E
r
 r 
1  
r 

1/2

G
mE
r
 r 
1 

2r 

mE
r3
(b)
K

1
m v2
2

K  K 
K  mv v  m G
U  G
mmE
r
U  G

mE r
r 2
1
1
2
m  v  v   mv 2  mvv
2
2
G
U  U  G
mmE
mE
G
r
3
2r 2
r
m mE
m mE  r 
 G
1  
r  r
r 
r 
m mE
r
r2
Energy work theorem:
W  K  U  
G mE m
r
2r 2
1