Download 1-9 Energy Homework

Name
_
AP Advanced Physics
Energy Written
1.
Problems
A 1200 kg elevator driven by an electric motor can safely carry a maximum load of 800 kg. What is
the power provided by the motor when the elevator ascends with a full load at a speed of 2.3 m/s?
(45kW)
t:::- ~
::
fV
(\'LCO 4-~D(:)(q.«'Xj..~)
-\:::
2.
-= \4$ PN \
----~
A small object of mass m slides without friction around a loop-the-Ioop apparatus as shown below. It
starts from rest at point A at a height h above the bottom of the loop.
(a) What is the minimum value of h (in terms of R) such that the object moves around the loop
without falling off at the top (point B)? (5RI2)
eli
o-.t
\
'2.
,1-1"
"I '"~'I'D::,".
"
hI'
1.t"
~:q;;'L~~~
"'"
__
u
1.
t.?l
'v\ :::-""~ -\-
~:::
L
y-
\J :
"
'Zoo,
-
(b) If h = 3R, compute the following quantities
horizontal diameter:
(i)
Velocity (2(grf5)
2- ~
~~;\
L."(" + ~
"l. ~
::: 2.'("'r
for the object at point C, which is at the end of a
0- ~"":. 7--j ""c..
'2('J:R
'12..-;:
(ii)
-jr)
radial ac~e.J:ration ~4g) and
0. yo
-= r
1-
l1. JY"') -:=
-6]
(""
(ii i)
tangential
~~
"(V\~
acceleration
(g)
"1'C1 -=
~ ~o.."'<:.
-=
\N
'(Y""\
0,......
rio..;,.
:C\\
3.
A ball of mass 0.5 kg is tied to a string of lengt~5-m~he
other end of the string is tied to a
rigid support. The ball is held so that the string makes an angle of 60° with the vertical, with the
string pulled taut. The ball is then released from rest.
2•.•..•
(a) What is the speed of the ball at the lowes,t point o{ its motion? Include a diagram in your
~
=*~"
T
solution.
.s¥-~rJ
2~~~~\=~
>v/.o. h
(4.43m/s)
\
-= ::;.<- rr."
V ~ f2-i-'1-",,~i
N\
__
-= J1. (el.i') (') -=
----.]
t LJ • 4-3 ?
(b) What is the tension in the string when the ball is at the lowest point of its motion? Include an
FBD in your solution. (9.8N)
m"2..
,-VV
-::
-
,
B
R
~
_") '2'(" -/ R J :::')
,,'l-
-\-
t1 (oJ: s?-v.c\
.r..
:z.
4. A certain spring is faund NOT to obey Hooke's law, but rather exerts a restoring force F(x) = - 40 x
- 9 x' if it is stretched orcompressed a distance x. The units of the numerical factors are such
that if x is in meters, then F will be in newtons.
(a) Calculate the potential energy function U(x) for this spring. Let u=o when x=O. (20x'+3x3)
r: _ - d \A.
t...4O'i - q)( '2.) d;<. == -
1\.»-
C
di
l~\,,*,'\"')
\ V"= LO'/..1.. -+ ~)(~]
(\<\Ic<.t~)
c\1A.
iy
-4 o~+ ~Yo1- -=- -- 6.y.,
J
c\.)I. -= -~
0)( _ ~ )(.'Z.)
_ w~
d. v..
_L_. ------'
\c-::: -
'l. - "5 'J.
\A
An object of mass~
IS attached to this spring, pulled a distance 1.2 m to the right an a
frictionless horizontal surface and released. What is the velocity of the object when it is 0.5 m to
the right of the
Q-\1;<=oequilibrium position?(4.37 m/s) \.1fl'.
tA..-='LaC 1.2.) --I 3l,o'2..)2 -= ~~.C\~l.f-S
(<;"<'?A(~ \)w.-\o.:d-".l
(1;)
c.\v...
1.
\A-=00~.S)
-,
;-;(.sy _
c;.?:>153'
::
2- ~ .u>O(1":S
~J
\ v-= 4.'~1
5. The patential energy function of a particle is
Xv.
1,
~+-__
-') 0
Cl
S
10
(-ICO0)
~ ('300'))('2;-(-\2))(.:',.
1-""
U(x)=-k:c+bx+c
2-50
2
where k = 300 N/m, b = -12.0 N/m' and c = -1000 J.
ULlI.)::\SO(
f
~;.
-\2)(
-IOI::>D
(a) Plot the function fram x = -5.00 m to x =+ 15.0 m. attach a copy of the graph to this packet.
1-z.'5~ (b) Assume the particle has a mass of 50 kg and is released from rest x = -5.0 m. Describe the
~cJ:>
subsequent motion of the particle. Where is the particle movingfastest? Slowest? Is it ever at
-\000
IS ;\1':>0
Detwt-tY'\
re;t?
-S-D
l'l\Q!(2-)
II
0 - S ~.
cUcx-eo..';,.{s.
ct\{)~:
ro
uecP'''St:S
~(.-e
\-00
\-o~l~ \'<\C(-eW~ \-t f'C'~. C,'N\~'f\
\A 't' l;o ¥-€"
=- ')
~I'""\
=)I.A..,'bc¥-t't
~W-t ~ 'i?-\ S
II
f 'I'nt<~W ~\C\\~
(c) Assume the particle is released from rest at x = +5.0m. Describe the subsequent motion of the (1A.li
particle.
~~et')
(d) If the particle is released at x = -5.0 m with a speed of 10 mis, how fast will it be going at x = 0.8
I,.L.
m? (17.5 m/s)
~\ ~()6
\ l.J'\1..oU<-
-'5 rY"
\
-= v..CJ + '(.. -=- '-12'50
It,1S0:) -= 1. rf'V
lP1C;O
7.. -t
:::1-5N'l.-\-
l _---J
vl':SOfq\O.\y'\
~'S
()L ') -;::-..L
(COO"" V"2. +
7-
9ty-
_
~.\Lj
-;:-:.JL.
v
( ~( 0)2..
-+ '2. 50) 1
1'5 D (..
-=:::.
q):L _
~
I SO J
11...'• '/)?> -I
\,;
'J
-1000
I
Ih'
V-::"\J'bOv.T
~S'"
=~
COt)
(j;)..
A tow truck is pulling a car of mass M=2000kg attached vdth a cable as
shown. There is also frictional force on tbe Cill. At position x=O the car bas
velocity
Vo = 5m/s.
~~@
z: __
%&0
a. In the space below
forces acting on it.
draw
the free body
~.~
diagram
for the car,
showing
all
~_L\
~rv-~
b.
What
is the kinetic
energy
KE of the car at x=O?
I
2..'5pOOT
IKE=
:~'J••..
1
If the total frictional force is a constant Ff ::: SON. what is the work Wf
done by friction on the car when it moves from x ::: 0 to x =150 m?
c.
I_
l,500:J
d.
"1-.',
The low truck pulls the cable
so that the tension
T in the cable
~J
300
lie
.s
changes. as shown in the graph.
200
What is the work WT done on the
~
car by the tension force when
a
=
the car moves
,,=150 m.
from
x=O to
~
100
o
W-==t=.X
\ 'SQOQ
e.
+
o
15'00
50
100 150 200
position x In m
IWT=22 1500:r
N." '" f:::,'.; €
2./L,SDO -l,'SOO -= ~
I
'{\l
-2S',UQoKE=
~.c:: /'5,000
I
.)..;;J:.SLj
What is the kinetic energy KE of the car at x = 150m?
i-fO,Ot)OT
+'2.$',000
f. What is t'he' power PT supplied to the car by the tension force
initial point x = 07
at the
-------
-
--
---_.-
A net force acting on a glider on an air track varies with its position x
along the track as shown in the graph. The glider, which has mass =
O.Bkg. moves without friction.
Force
in N
6
o
2
x in m
-6
(b) If the velocity of the glider is zero at x = O. find the kinetic energy K and
tile velocity v of the glider at the position x = 1m.
\f'J -::
tX-te
:::'3 -= ~t:'- 0
2.( 31')-= l~~ V'2.j 2
r..o -= (-.-t) V '"l~N{
\/'1-=
~
I
_K=_;,_j
I
~' ' G T)
I---!O.-l
=)\j::J{ t]. .-,t-t s (i. ht1ti~)
v=
(c) Suppose now that the initial velocity of the glider at x:::: 0 is 0.5 mls. The
same force shown in the graph above the acts on the glider. What are the
kinetic energ.v K and the velocity v of the glider at the position x = 1m.
0'\
[) )
A mass m is lied to a string of length L. The mass is released
{rom rest at a height h as shown in the figure.
Give your answers
below as simple expressions
in terms of the symbols defined and the
acceleration
of gravity g.
P
e_--"!
I
'"
1(;\
-~O
B
3.
Find the speed VB of the mass at the bottom
marked B on the figure.
~~
of its trajectory,
~ ~ rf-v'-
= 2..3 h
. \J 2.
, Y::.J~Cjn'
b. When the mass is at the point B. -the string hits a
is a distance d above B. The mass continues around
d to the point P where the string is horizonlaI.
Find
the mass at point P.
¥-c 'f. E t'
i. rf rt-S '<\) ~ 2
~
II
.
v;}
?
1
!
2"
-='..J
CJ\.tJ
!
-=
Ii gu re.
D--y- :: Cl;
V
vp
~d
=
J 2(3""' -<j d)
1
1Jj (Y, -,,)
"2-
P
'l-\ -
l ~#- ...lr 0..\
-'<l
c\
d
z~~~
;''r#.~...
'd.' \Vh;{
I
c;\)
,_V 7- _ '2:..:.J0->-(~_- ...••
o...0=-C4
A -:.JIlI.1-{().('V\,
.1..
-\
rf'jq
~
c.
at is the acceleration
vector ap of the mass al the point P?
Give your answ~r in terms of the unit vectors I and j shown in the
vY'~~c:\("
o
J CI-"2.
SI;
~T\
:i Vp"L
~
'- -
t> <.----.-
I,
'v'I
Y'j\ Vrr'-
~d\
'?:.( ~'" V -:::.J 1..l5"" - ~ d)
If' '2. =:
T U ':lp
(?" -;::
t
fixed rod which
a circle of radius
the speed vp of
,.
is [he tension
TI, l.'3CY--d)1--+;'I .. ----:--:c:..=J' -d
Tt'zt" y
''
J
T in the sIring when the mass is at point
;s. ~~~
__ 0 -:1
'l-I
T-:;'(Y\~q(h-dD
T=
r~j:'
P?
-:UY'O}_ v-..-o.')
-
d
.,
/j~
Q
\]JWr-<'
V-=
0~-dV2
J2~Ch-d)
'2'jC'"'-<0-
C;-O\\l e -0>(
d
VJ
-\1>
-------
f1,
-I ) .
A skier starts from rest at th(>top ora ski jump as sho •••.
'll. (Neglect Friclion.}
A
h B= 10
~r--
::::r:e =
C
B
15'
'
-----r---
he: ::'5m
Ground Level
If the speed of the skier at point C is 20 mls. what is the speed
3.
u:ihe track. (B)
Tott"'" 'f.c
.l-if
'2.
('2.0)7-
-::¥-\b 1-
1- ~t
+?"q (1'5)
2-00 -+ 1'1' -
-:: ~
'H' -=
VB
at the bottom
v,g
i~
VEo~
~
+ ';0a(lo)
'5'l.
J
b. \\1':2t is the starting height h,\?
rr< ~'" "" :=
''t\p.. -=-
'rf\ l'2..t>
~
2.P 0
y- .HY\ Ij 15
-'r \'11
I
[h =3'5.4YY")
"I,r
A
:5 s.. '1 i},?
c. At point C the track ends v."'ith a slope of 15C• How high v.ill tl:e skier
rise abo'!:€'
troe ground after leaving the track?
mfA-X
\If
~%~
V'1r,.= 0
V'j:-::
1-: v ( 'l-+ '2"3 t:>)l.
( c; i\~)
::::bY.. = \.~..,"" -I- ISm
'2..(0..«\
d.
tt the mass
I
hm •• =
I~.4/V1
I
110, 'j rJ1
of the skier is 75 Kg. and the radius of curvature of :.he traCk at B i~
R=30 m. what is force F orthe track on the skier at point B?
IF= \q lKN I
I 'i -f'>.;
----
.--
_---
•.
I\.i