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physics 1 lab: newton's second law

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Course
Physics 1 Lab (PHYS1511)

Institution
Fordham University

The notes for the physics lab on Newton's Second Law focus on verifying the relationship between force, mass, and acceleration, expressed as

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Uploaded on December 26, 2024
Number of pages 3
Written in 2023/2024
Type Class notes
Professor(s) Holler
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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

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