physics 1 lab: newton's second law

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