N
A
R
E
W
S
S
Suspended Accelerating-Objects
A resultant force causes a system to accelerate. The direction of the acceleration is in the direction of the resultant
force. As illustrated below, when suspended objects accelerate, they do so in the direction of the resultant force.
Stationary (at rest)
(w = T) FR = 0
Accelerating upward
(T > w) FR = T - w
T
Going
up
T
w
Accelerating downward
(w > T) FR = w - T
T
FR = m a
T - w = ma
-w
a = Tm
w
w
Coming
down
FR = m a
w - T = ma
-T
a = wm
Frictionless Pulley
When two masses are attached at opposite ends of a “light” cord that passes over a frictionless pulley, the two
masses act as a single mass whose acceleration is as follows:
m2
m1
 w1  w 2
FR  w 2  w 1
 FR  ma
FR  0
a0
m1
m1
 w 2  w1
m2
T
m1
T
m2
a
w 2  w1
m1  m 2
FR  w1  w 2
T
m1
w1
w2
T
 FR  ma
T
m2
Total mass
w1
m2
 w1  w 2
a
w1  w 2
m1  m 2
T
m1
m2
w2
Total mass
w2
w1
 Note: Use 10 m/s2 for the acceleration due to gravity.
1.
A light (i.e. massless) cord passes over a pulley that turns on a frictionless
axis as illustrated in the diagram on the right. Two masses, 3 kg and 2.6 kg,
are attached at opposite ends of the cord. Determine the magnitude of the
acceleration of the masses. [0.7 m/s2]
2.6 kg
3 kg
Note that the cord unites the objects int o a sin gle mass of 5.6 kg.
____________________________________________________________
FR  w 1  w 2  m 1 g  m 2 g  ( 3 kg)(10 m / s 2 )  ( 2.6 kg)(10 m / s 2 )  4 N
_______________________________________________________________________
 FR  ma
_______________________________________________________________________
F
4N
 a R 
 0.7 m / s 2
_______________________________________________________________________
m 5.6 kg
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC
© 1999 S. Lancione
Page 1 /8
2.
A stone is mechanically
v i  30 m / s shot straight up in the air with a velocity of 30 m/s.
v f does
0 the stone rise? [45 m]
a) How high
v  v i 0  30 m / s
_____________________________________________________________________
va  f

 15 m / s
s  va t
2
2
_____________________________________________________________________
 (15 m / s )( 3 s )
_____________________________________________________________________
 45 m
_____________________________________________________________________
1
s  v i t  at
b) At what time will2 the stone be 40 m from the ground on its way up?
or 40  30 t  5t 2
[2 s]
_____________________________________________________________________
2
N
o
t
e
:
T
h
e
u
n
i
t
s
a
r
e
n
o
t
i
n
c
l
u
d
e
d
t
o
s
i
m
p
l
y
5 t  6t  8  0
t
h
e
m
a
t
h
e
m
a
t
i
c
a
l
e
x
p
r
e
s
s
i
o
n
s
.
_____________________________________________________________________
(t  2)(t  4)  0
_____________________________________________________________________
 t  2 s and t  4 s
Answer : t  2 s (on its way up)
_____________________________________________________________________
c) At what time will the stone be 40 m from the ground on its way down? [4 s]
 t  2 s and _____________________________________________________________________
t 4s
1
N
o
t
e
:
T
h
e
u
n
i
t
s
a
r
e
n
o
t
i
n
c
l
u
d
e
d
t
o
s
i
m
p
l
y
s  vdown
a)t
it 
Answer : t  4 s (on its way
t
h
e
m
a
t
h
e
m
a
t
i
c
a
l
e
x
p
r
e
s
s
i
o
n
s
.
2
_____________________________________________________________________
or 40  30 t  5t 2
_____________________________________________________________________
2
5 t  6t  8  0
_____________________________________________________________________
( t  2)( t  4)  0
3.
A mass of 20 kg hangs by a rope from a ceiling inside an elevator.
a) What will be the tension in the rope if the elevator
is accelerating upward at 4 m/s2 ? [280 N]
T
20 kg
_____________________________________________
S ince the elevator is going up : FR  T  w
_____________________________________________
 FR  ma
a = 4 m/s2
W
 T  w  m a or T  mg  ma
_____________________________________________
T  ma  mg  ( 20 kg)(4 m / s 2 )  ( 20 kg)(10 m / s 2 )  280 N
_____________________________________________
b) What will be the tension in the rope if the elevator
is accelerating downward at 4 m/s2 ? [120 N]
_____________________________________________
S ince the elevator is going down : FR  w  T
_____________________________________________
 FR  ma
T
20 kg
a = 4 m/s2
W
 w  T  m a or mg  T  ma
_____________________________________________
T  mg  ma  ( 20 kg)(10 m / s 2 )  ( 20 kg)(4 m / s 2 )  120 N
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 2 /8
4.
A stone is thrown vertically upward with a velocity of 15 m/s from the bottom of a cliff
that is 15 m high. At the same time, another stone is dropped from the top of the cliff.
At what height from the ground will the two stones meet? [10 m]
s Down  s up  15 m
_______________________________________________________________________
1
N
o
t
e
:
T
h
e
u
n
i
t
s
a
r
e
n
o
t
i
n
c
l
u
d
e
d
t
o
s
i
m
p
l
y
But s  v i t  at 2
t
h
e
m
a
t
h
e
m
a
t
i
c
a
l
e
x
p
r
e
s
s
i
o
n
s
.
_______________________________________________________________________
2
1
1
_______________________________________________________________________
 v Down t  at 2  v Up t  at 2  15 m
2
2
1
_______________________________________________________________________
 s  v i t  at 2
2
2
or 0  5t  15t  5t  15
2
_______________________________________________________________________
1
 15t  15
 0  (10)(1) 2  5 m
_______________________________________________________________________
2
t 1s
Thus
,
the
ball
falls 5 m after 1 s.
_______________________________________________________________________
Answer : 15 m  5 m  10 m
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
5.
A boy drops a stone vertically downward from a bridge 125 m high. One second later,
he throws another stone also vertically downward. With what initial velocity must he
throw the second stone so that both stones strike the water at the same time? [11.3 m/s]
_______________________________________________________________________
Note that while the dis tan ce for both stones is the same (125 m), the times are
_______________________________________________________________________
not the same. If t represents the time of fall for the first stone, then the time
_______________________________________________________________________
for the sec ond stone is ( t  1) sin ce it was thrown one sec ond later.
_______________________________________________________________________
Initial velocity of second stone
Time of first stone
_______________________________________________________________________
1
1
_______________________________________________________________________
s  v i t  at 2
s  v i t  at 2
2
2
_______________________________________________________________________
1
1
125 m  v i (4 s )  (10 m / s 2 )(4 s ) 2
125 m  0  (10 m / s 2 )(t 2 )
2
_______________________________________________________________________
2
2
2
125 m  v i (4 s )  (5 m / s 2 )(16 s 2 )
125 s  5t
_______________________________________________________________________
4 v i  45 m / s
or
t 2  25 s 2
_______________________________________________________________________

v i  11.25 m / s or 11.3 m / s

t  5s
_______________________________________________________________________
Time of second stone
_______________________________________________________________________
t  5s 1s  4s
_______________________________________________________________________
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 3 /8
6.
A rock is thrown vertically down with a velocity of 25 m/s. With what velocity will it
strike the ground 100 m below? [51 m/s]
_______________________________________________________________________
_______________________________________________________________________
 2a s  v f2  vi2
_______________________________________________________________________
 v f2  2a s  vi2
or v f2  2(10 m / s 2 )(100 m)  (25 m / s) 2
_______________________________________________________________________
or v f2  2625 m 2 / s 2
_______________________________________________________________________
 v f  51.2 m / s  51 m / s
_______________________________________________________________________
_______________________________________________________________________
7.
A pebble is dropped from the roof of a high building. If it takes 1.5 s to travel the last 100 m,
how high is the building? [275.2 m]
_______________________________________________________________________
Calculatio n of initial velocity for last 100 m.
_______________________________________________________________________
1 2
1
2
2
s  vi t  at
or 100 m  v i (1.5 s)  (10 m / s )(1.5 s)
2
2
_______________________________________________________________________
2
2
or
100 m  (1.5 s) v i  (5 m / s )(1.5 s)
_______________________________________________________________________
or v i (1.5 s)  11.25 m  100 m
88.75 m
_______________________________________________________________________
 vi 
 59.16 m / s  59.2 m / s
1. 5 s
_______________________________________________________________________
Calculatio n of height of top part of building
_______________________________________________________________________
2
2

2a s  v f  v i
_______________________________________________________________________
v f2  v i2 (59.2 m / s) 2  0 3504.64 m 2 / s 2



 175.2 m
2a
2(10 m / s 2
20 m / s 2
Thus, the height of the building is : 100 m  175.2 m  275.2 m
8.
s
With what initial upward velocity must a ball be thrown in order to rise 20 m? [20 m/s]
_______________________________________________________________________
 2a s  v f2  v i2
_______________________________________________________________________
v i2  v f2  2as
_______________________________________________________________________
v i2  0  2(10 m / s 2 )(20 m)
_______________________________________________________________________
v i2  400 m 2 / s 2
_______________________________________________________________________
 v i  20 m / s
_______________________________________________________________________
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 4 /8
9.
A package is attached to a helicopter ready to be released (dropped). If the helicopter is
ascending at 5 m/s and releases the package from an altitude of 550 m, how long does it
take the package to strike the ground? [11 s]
N
o
t
e
:
T
h
e
u
n
i
t
s
a
r
e
n
o
t
i
n
c
l
u
d
e
d
t
o
s
i
m
p
l
y
t
h
e
m
a
t
h
e
m
a
t
i
c
a
l
e
x
p
r
e
s
s
i
o
n
s
.
_______________________________________________________________________
Since the helicopter is ascending ,
U sin g the quadratic equation
_______________________________________________________________________
the accelerati on of the package
is negative when released .
_______________________________________________________________________
 B  B2  4A C
1 2
t

_______________________________________________________________________
 s  vi t  a t
2A
2
_______________________________________________________________________
1
 (1)  (1) 2  4(1)(110)
550  (5 ) t  (10) t 2
t

2
_______________________________________________________________________
2(1)
2
or 550   5t  5t
_______________________________________________________________________
1  1  440 1  441 1  21



or t 2  t  110  0
2
2
2
_______________________________________________________________________
 A  1 B   1 C   110
1  21 22


 11 s (discard negative value)
_______________________________________________________________________
2
2
10.
Starting from rest, a block of wood takes 5 s to slide down an inclined plane 120 cm long.
a) What was its acceleration?
[0.1 m/s2]
____________________________________________________________________
____________________________________________________________________
1
 s  vi t  a t 2
____________________________________________________________________
2
1
____________________________________________________________________
or 1.2 m  0  (a )(5 s) 2
2
____________________________________________________________________
2(1.2 m)
 a
 0.096 m / s 2  0.1 m / s 2
2
2
25 m / s
____________________________________________________________________
____________________________________________________________________
b) With what speed did it reach the bottom of the incline?
[0.5 m/s]
____________________________________________________________________
____________________________________________________________________
 2a s  v f2  vi2
____________________________________________________________________
or v f2  vi2  2a s  0  2(0.1 m / s 2 )(1.2 m)  0.24 m 2 / s 2
____________________________________________________________________
 v f  0.489 m / s  0.5 m / s
____________________________________________________________________
____________________________________________________________________
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 5 /8
11.
A ball falls from rest and is seen to pass a window 2 m high in 0.1 s.
Determine the height above the window that the stone fell? [19 m]
________________________________________________________
Initial velocity of ball upon pas sin g top of window :
V=0
h=?
V = 19.5 m/s
1
________________________________________________________
 s  vi t  a t 2
2
________________________________________________________
s 1 a t 2 s a t 2 m (10 m / s 2 )(0.1 s)
2m

v


 


 19.5 m / s
i
________________________________________________________
t 2 t
t 2 0.1 s
2
_______________________________________________________________________
Final velocity of ball upon pas sin g bottom of window :
v
_______________________________________________________________________
V = 20.5 m/s
 a
  v  a t  (10 m / s 2 )(0.1 s)  1 m / s
t
_______________________________________________________________________
But  v  v f  v i
_______________________________________________________________________
 v f   v  v i  1 m / s  19.5 m / s  20.5 m / s
_______________________________________________________________________
Total dis tan ce ball falls :
_______________________________________________________________________
 2a s  v f2  v i2
_______________________________________________________________________
v 2  v i2 (20.5 m / s) 2  0
420.25 m 2 / s 2
 s f


 21.0 m
_______________________________________________________________________
2a
2(10 m / s 2 )
20 m / s 2
 Dis tan ce above window  Total dis tan ce  Height of window
_______________________________________________________________________
 Dis tan ce above window  21.0 m  2.0 m  19 m
12.
Two stones, A and B, are respectively 45 m and 60 m above the ground.
If stone A is dropped at the same time as stone B is thrown vertically
down with a velocity of 12 m/s, which stone hits the ground first? [B]
_________________________________________________________
B
12 m/s
A
N
o
t
e
:
T
h
e
u
n
i
t
s
a
r
e
n
o
t
i
n
c
l
u
d
e
d
t
o
s
i
m
p
l
y
Time stone  A strikes ground
t
h
e
m
a
t
h
e
m
a
t
i
c
a
l
e
x
p
r
e
s
s
i
o
n
.
_________________________________________________________
1
 s  vi t  a t
_________________________________________________________
2
U sin g the quadratic equation
(
10
m / s2 )t 2
_________________________________________________________
45 m  0 
2
 B  B 2  4A C
_________________________________________________________
t

2
(
45
m
)
or t 2 
 9 s2
2A
2
10
m
/
s
_________________________________________________________
2
45 m
60 m

t  3s
 12  (12) 2  4(5)(60)
_______________________________________________________________________
t
2(5)
Time stone  B strikes ground
_______________________________________________________________________
1
 12  144  1200  12  1344
 s  vi t  a t 2


_______________________________________________________________________
2
10
10
or 60 m  (12 m / s) t  (5 m / s 2 ) t 2
_______________________________________________________________________
 12  36.6 24.6


 2.46 s  2.5 s
or 5 t 2  12 t  60  0
10
10
_______________________________________________________________________
 A  5 B  12 C   60
(discard negative value)
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 6 /8
13.
Two masses, 8 kg and 2 kg, are connected by a light rope that
passes over a frictionless pulley (see diagram). Find:
a) The magnitude of the acceleration of the masses. [2 m/s2]
FN = 80 N
8 kg
___________________________________________________________________T
2 kg
Fw = 80 N
___________________________________________________________________
 F  ma
R
F
20 N
___________________________________________________________________
W
 a R 
 2 m / s2
m 10 kg
___________________________________________________________________
___________________________________________________________________
b) The tension in the cord. [16 N]
___________________________________________________________________
FR  ma
___________________________________________________________________
FR  20 N  T
___________________________________________________________________
 ma  20 N  T
___________________________________________________________________
T  20 N  ma  20 N  ( 2kg)( 2 m / s 2 )  16 N
___________________________________________________________________
14.
Two masses, 2 kg and 8 kg, are connected by a light rope that
passes over a frictionless pulley (see diagram). Find:
a) The magnitude of the acceleration of the masses. [8 m/s2]
FN = 20 N
2 kg
T
8 kg
___________________________________________________________________
Fw = 20 N
___________________________________________________________________
 FR  ma
F
80 N
___________________________________________________________________
 a R 
 8 m / s2
W
m 10 kg
___________________________________________________________________
___________________________________________________________________
b) The tension in the cord. [16 N]
___________________________________________________________________
FR  ma
___________________________________________________________________
FR  80 N  T
___________________________________________________________________
 ma  80 N  T
___________________________________________________________________
T  80 N  ma  80 N  (8 kg)(8 m / s 2 )  16 N
PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 7 /8
15.
A 60 kg person stands on a spring scale in an elevator.
If the elevator is accelerating upwards at 2 m/s2, what
is the reading on the balance? [720 N]
3
4
5
2
6
1
7
____________________________________________
____________________________________________
FR  ma
F
But FR  F  w
____________________________________________
 F  w  ma
______________________________________________________________________
F  600 N  (60 kg)( 2 m / s 2 )
______________________________________________________________________
W
F  600 N  120 N
______________________________________________________________________
 F  720 N
FR = F - w
______________________________________________________________________
75
Balance reading
______________________________________________________________________

PHYSICS 534  DYNAMICS  AnsPhysics Ex-31.DOC  Free Fall Part-2 /2
Page 8 /8