lab 5 : newton's second law
·
newton's second law ·
question 2 ·
question 3 d
·
the sum of all forces on an object is equal to mass t i m e s acceleration
F
& A A
O ·
acceleration
F = ma
* direction of Fym = ma =
mg -
T
I left block
question the
mg-md
·
T =
k
2kg Gkg 79
2k9dg
Sky 319 419 lg ,
free-body diagram of box FxM Ma = T
·
=
A B C E
T =
Ma
&
T Fm ,
= m , a = m , G -
T
= Ma
=
Fm = mzd =
T mg-ma
m2g
-
z
20kg a = + 4 9 m/s
.
a = mg
m g m a mid m2g
= +
, ,
-
M + m
- mg
a =
g(m -ma)
don't need
asking for I al
,
since its we ↑
question I d question be
· ·
m
, + m2 to rewrite the formula
·
determine the mass of the block free-body diagram
A: 2 . 45 M/s -Nd
T
Fbox
,
=
mpdb = T mpg Es) < A < B <D
27 m/s
-
B : 3
d
.
T
5k M -
T =
mpd +
mbg
= 294N
c : 0 754 M/s
m
.
&
D : 4 20 m/s -
F
.
Block = mpd =
mB9 -
T mg
E : Om/s same weight in equilibrium -mg
T =
mB(g -
a)
question 3 question 3f
=T CAN O
mB
· ·
=
free-body diagram acceleration
·
question 1b - do Fym = ma =
mg -
T
N T
T mg-md
free body
=
diagram of block a acceleration of system d
-
M D T
now the block moves up bic Wbox > Wblock m
F
xM
=
Ma =
T -
MkN
*
T
T
N =
Mg
Fbox
g
= mpd =
mpg -
MG -mg
60kg -d T
T =
Ma + uMg
=
mpg -
mad
question 3 d
·
mg Ma
-
F
block =
mid = T -
mBg mg-ma = +
uMg
·
acceleration a tension of blocks
T =
mid +
mBG a = g(m -
uM)
my =
100k9 assuming that the mass a friction of the string is negligible
M + m
& the string does not stretch the acceleration a tension of
mB =
60kg mBd + mBg =
mbg -
mbd
,
the blocks are equal in magnitude bic they are connected by
·
procedure 1
a =
g(my -
mB) =
2 . 45 m/s a
3 M/s the same string
what happens to the velocity if there was no friction
(MB + mp) ( + ) for block * tension remains constant throughout the string
(-) for box ·
question 3b thevelocityofthecartwouncrease moreila ,
·
equation for m a ss m
question 1C
·
·
procedure 4
free body diagram of block a acceleration of system Fy =
ma =
mg -
T
free-body diagram a find MK
·
n ow the block moves down blc Wbox < Wblock
T
m =
and
,
a 9
*
T Fx fk MkN
-
=
ma = =
Fbox =
mpd = T -
mbg
N =
mg
question 3C ·
·
135k9 d T =
mpd +
mbg
,
fi
equation for m a ss M
g ma
ukmg
·
=
Fblock =
mBd =
mBg -
T
- mg ma
Mk = = d
Fx =
Ma = T
T =
mBg -
mBd mg g
M
T
=
mBg mBd mbg procedure S
=
mpd +
-
·
·
what does the th coefficient m e a n ?
a = g(mB -
mb) = 1 . 46 m / s a
2 m/s
0 = a + bt + c t2
mb + MB ( + ) for box A
·
Crepresents a bic the quadratic is
(-) for block
analogous to x =
Xo + vt + at
D
time
·
newton's second law ·
question 2 ·
question 3 d
·
the sum of all forces on an object is equal to mass t i m e s acceleration
F
& A A
O ·
acceleration
F = ma
* direction of Fym = ma =
mg -
T
I left block
question the
mg-md
·
T =
k
2kg Gkg 79
2k9dg
Sky 319 419 lg ,
free-body diagram of box FxM Ma = T
·
=
A B C E
T =
Ma
&
T Fm ,
= m , a = m , G -
T
= Ma
=
Fm = mzd =
T mg-ma
m2g
-
z
20kg a = + 4 9 m/s
.
a = mg
m g m a mid m2g
= +
, ,
-
M + m
- mg
a =
g(m -ma)
don't need
asking for I al
,
since its we ↑
question I d question be
· ·
m
, + m2 to rewrite the formula
·
determine the mass of the block free-body diagram
A: 2 . 45 M/s -Nd
T
Fbox
,
=
mpdb = T mpg Es) < A < B <D
27 m/s
-
B : 3
d
.
T
5k M -
T =
mpd +
mbg
= 294N
c : 0 754 M/s
m
.
&
D : 4 20 m/s -
F
.
Block = mpd =
mB9 -
T mg
E : Om/s same weight in equilibrium -mg
T =
mB(g -
a)
question 3 question 3f
=T CAN O
mB
· ·
=
free-body diagram acceleration
·
question 1b - do Fym = ma =
mg -
T
N T
T mg-md
free body
=
diagram of block a acceleration of system d
-
M D T
now the block moves up bic Wbox > Wblock m
F
xM
=
Ma =
T -
MkN
*
T
T
N =
Mg
Fbox
g
= mpd =
mpg -
MG -mg
60kg -d T
T =
Ma + uMg
=
mpg -
mad
question 3 d
·
mg Ma
-
F
block =
mid = T -
mBg mg-ma = +
uMg
·
acceleration a tension of blocks
T =
mid +
mBG a = g(m -
uM)
my =
100k9 assuming that the mass a friction of the string is negligible
M + m
& the string does not stretch the acceleration a tension of
mB =
60kg mBd + mBg =
mbg -
mbd
,
the blocks are equal in magnitude bic they are connected by
·
procedure 1
a =
g(my -
mB) =
2 . 45 m/s a
3 M/s the same string
what happens to the velocity if there was no friction
(MB + mp) ( + ) for block * tension remains constant throughout the string
(-) for box ·
question 3b thevelocityofthecartwouncrease moreila ,
·
equation for m a ss m
question 1C
·
·
procedure 4
free body diagram of block a acceleration of system Fy =
ma =
mg -
T
free-body diagram a find MK
·
n ow the block moves down blc Wbox < Wblock
T
m =
and
,
a 9
*
T Fx fk MkN
-
=
ma = =
Fbox =
mpd = T -
mbg
N =
mg
question 3C ·
·
135k9 d T =
mpd +
mbg
,
fi
equation for m a ss M
g ma
ukmg
·
=
Fblock =
mBd =
mBg -
T
- mg ma
Mk = = d
Fx =
Ma = T
T =
mBg -
mBd mg g
M
T
=
mBg mBd mbg procedure S
=
mpd +
-
·
·
what does the th coefficient m e a n ?
a = g(mB -
mb) = 1 . 46 m / s a
2 m/s
0 = a + bt + c t2
mb + MB ( + ) for box A
·
Crepresents a bic the quadratic is
(-) for block
analogous to x =
Xo + vt + at
D
time
