lab 7 : range prediction
translational kinetic energy procedure 3
rotational kinetic energy
· ·
·
exit speed ?
k mu energy stored in the spinning / rotating of an object
·
·
=
when a ball rolls down a ramp we need to consider translational d rotational kinetic energy
Imusi mus mani
·
gravitational potential
·
energy Inve +
Imue might =
+ + wi
ug =
migh
Kr =
- mu
total mechanical question 4a
to mve
·
might
energy
·
+ =
mgho + We
of kinetic a potential
·
find vea Ave
sum
energy
·
·
law of conservation of mechanical energy k+ + kr +
Ug
ve =
(g(no -h + ) + We
·
total mechanical
·
the total amt of energy is constant a conserved
energy cannot be created nor destroyed
Imie +
Inve might =
Imusimuomeni Ve =
1 g (no -h + + n +b -
hib)
[muf might Emvo maho
tomve
+ =
+ ?
=
mg (ni - h+ * time of flight ?
·
when is total mechanical not conserved
·
when non-conservative forces are present energy is not conserved
Pg(hi -h )
,
ve =
+
friction a tension cannot be conserved nor converted into another form of
energy
·
question 3 a
·
·
what is the speed ve ?
19 .
2Ah
gah
Ave = I
O
2
g(hi h +)
imvo Inve ↳ g(ni
-
+ mgho = +
might -
n+
a range ?
starts from rest
* question 4b
Inve =
mgho-might
time of flight of the ball & DT
·
ve = 29(no -h +)
question 3b
·
procedure 4
·
uncertainty of ve equation
·
Ave ?
D2g(no -
hH) =
2g
.
2Ah
range of the ball & AR !
·
Ve
& (no hib)
=
g -h + + n +b
49Ah 29Ah
-
=
a 29(ho-hH =
- 29(no-hi zg(no-h +
-
29Ah 19 .
4ah
Ave =
2g(no h +) conservation of wh friction
·
energy
-
there i s friction i n the tube that causes the ball to lose
AVe =
19 . 49h
energy as it goes from hi to he
·
question 3C
·
2
·
time of flight ,
T ,
how long the block stays in the air before landing
Ki +
Ugi + Nfriction =
Kf +
Ugf (g(no-hf + h + b-hib)
↳ K : total KE both translational d rotational
sy = vot i Eat ,
procedure 1
·
X = h+ , a =
g
·
equation for We
t = 2hf
9
[mv +
Imvo +
mghin +
We
=
EmV +
5 mv + +
mghtb
·
At ?
starts at rest ends at rest
2Ah +
Dt = g Ahe1g
=
We =
mg (h + - hib)
1
2h+g Ch + 19
·
procedure 2
question 3d *
·
·
would equation from We be valid if him d htb differed ?
·
determine the range
since the work done by friction is non conservative work ,
it depends on the force of friction times the
R =
Vet +
zat2 displacement of the object . Therefore , We is not dependent on the distance traveled /path taken
,
range is i n the x-direction so a = 0 blc V is constant but rather ,
the displacement . bic the equation only considers the displacement through Ah ,
the
expression remains valid no matter the heights a where the ball is launched from
Ch
R =
Ve .
t =
29(ho-hf) .
AR =
BVe + Ve .
T = AVeVe +A
translational kinetic energy procedure 3
rotational kinetic energy
· ·
·
exit speed ?
k mu energy stored in the spinning / rotating of an object
·
·
=
when a ball rolls down a ramp we need to consider translational d rotational kinetic energy
Imusi mus mani
·
gravitational potential
·
energy Inve +
Imue might =
+ + wi
ug =
migh
Kr =
- mu
total mechanical question 4a
to mve
·
might
energy
·
+ =
mgho + We
of kinetic a potential
·
find vea Ave
sum
energy
·
·
law of conservation of mechanical energy k+ + kr +
Ug
ve =
(g(no -h + ) + We
·
total mechanical
·
the total amt of energy is constant a conserved
energy cannot be created nor destroyed
Imie +
Inve might =
Imusimuomeni Ve =
1 g (no -h + + n +b -
hib)
[muf might Emvo maho
tomve
+ =
+ ?
=
mg (ni - h+ * time of flight ?
·
when is total mechanical not conserved
·
when non-conservative forces are present energy is not conserved
Pg(hi -h )
,
ve =
+
friction a tension cannot be conserved nor converted into another form of
energy
·
question 3 a
·
·
what is the speed ve ?
19 .
2Ah
gah
Ave = I
O
2
g(hi h +)
imvo Inve ↳ g(ni
-
+ mgho = +
might -
n+
a range ?
starts from rest
* question 4b
Inve =
mgho-might
time of flight of the ball & DT
·
ve = 29(no -h +)
question 3b
·
procedure 4
·
uncertainty of ve equation
·
Ave ?
D2g(no -
hH) =
2g
.
2Ah
range of the ball & AR !
·
Ve
& (no hib)
=
g -h + + n +b
49Ah 29Ah
-
=
a 29(ho-hH =
- 29(no-hi zg(no-h +
-
29Ah 19 .
4ah
Ave =
2g(no h +) conservation of wh friction
·
energy
-
there i s friction i n the tube that causes the ball to lose
AVe =
19 . 49h
energy as it goes from hi to he
·
question 3C
·
2
·
time of flight ,
T ,
how long the block stays in the air before landing
Ki +
Ugi + Nfriction =
Kf +
Ugf (g(no-hf + h + b-hib)
↳ K : total KE both translational d rotational
sy = vot i Eat ,
procedure 1
·
X = h+ , a =
g
·
equation for We
t = 2hf
9
[mv +
Imvo +
mghin +
We
=
EmV +
5 mv + +
mghtb
·
At ?
starts at rest ends at rest
2Ah +
Dt = g Ahe1g
=
We =
mg (h + - hib)
1
2h+g Ch + 19
·
procedure 2
question 3d *
·
·
would equation from We be valid if him d htb differed ?
·
determine the range
since the work done by friction is non conservative work ,
it depends on the force of friction times the
R =
Vet +
zat2 displacement of the object . Therefore , We is not dependent on the distance traveled /path taken
,
range is i n the x-direction so a = 0 blc V is constant but rather ,
the displacement . bic the equation only considers the displacement through Ah ,
the
expression remains valid no matter the heights a where the ball is launched from
Ch
R =
Ve .
t =
29(ho-hf) .
AR =
BVe + Ve .
T = AVeVe +A
