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PHY121
Ch 6-9 Exam
Name
You are a member of an alpine rescue team and must get a 3.00
kg box of supplies up a 30° incline so that it reaches a stranded
skier who is a vertical distance 4.00 m above the bottom of the
incline. There is some friction present, µ = 0.100. Since you can't
walk up the incline, you give the box a push that gives it an initial
velocity; then the box slides up the incline, slowing down under
the forces of friction and gravity. To aid in the ascent, you
attach one of your bottle rockets to the back side of the box of
supplies, as shown, which exerts a force of 10.0 N. Use the
work-energy theorem to calculate the minimum speed v that you
must give the box at the bottom of the incline so that it will
reach the skier.
TEbottom
=
TEtop
[KEpush + PEbottom] + FR-in dir d –
Ff d
= KEtop + PEtop
KEpush + PEbottom + FR-in dir d
–
Ff d
= 0 + mghtop
½3v2 + 0 + 10 cos30° 8
µ FN 8
= m g htop
2
½3v + 0 + 69.3
µ FN 8
= 3 10 (4)
5/5
3/3
5/5
5/5
5/5
5/5
Note, please account for energy
FR d in
the initial energy, similar but opposite to
the non-conservative force, Friction
contributed by the bottle rocket,
through a distance,
5/5
2
1.5v
+ 69.3 - (0.1)(10sin30 + mgcos30) 8
v = 7.1 m/s
=
Ff d.
sin θ = h / d
sin 30 = 4 / d
d = 8 meters 1/1
120 J
A 50.0 kg roller-coaster car is released from rest and slides along a
frictionless track. At ground level, the car encounters a loop of radius R
= 10 m. If the track exerts a normal force of 625 N on the car at the
top of the loop, what is the initial height of the car?
Both mg and the normal force are toward the center, thus additive.
2/2
8/8
2/2
7/7
mg + 700 = mv2 / r
50(10) + 625 = 50 v2 / 10
50(10) + 625 = 50 v2 / 10
v =15 m/s
4/4
8/8
2/2
7/7
2
mg + 200 = mv / r
50(10) + 200 = 50 v2 / R
=
mghtrack + ½mvtrack
500(htrack) + 0
htrack = 31.25 m
A 50.0 kg roller-coaster car is
moving 4 m/s at h = 30 meters on
the top of the track and slides
along a frictionless track. At
ground level, the car encounters a
loop of radius, R. If the track
exerts a normal force of 200 N on
the car at the top of the loop,
what is R?
2/2
TEtoptrack
2
TEtoploop
2/2
= mghloop + ½mvloop2
= 500(2R) + ½50 152
OR for 80% max 
Both mg and the normal
force are toward the center,
thus additive.
4/4
8/8
TEtoptrack
A 50.0 kg roller-coaster car is
moving 10 m/s at h = 30 meters
on the top of the track and slides
along a frictionless track. At
ground level, the car encounters a
loop of radius, R = 10 m. What is
the normal force exerted by the
track on the car at the top of the
loop?
=
TEtoploop
2/2
mghtrack + ½mvtrack
2/2
2
= mghloop
4/4
+ ½ m vloop2
14 R = vtoploop2
4/4
TEtoptrack
=
2/2
mghtrack + ½mvtrack2
500(30) + ½50 42
15400
R = 11.4 meters
2/2
TEtoploop
500(30) + ½50 102
vloop = 17.3 m/s
= 500(2R) + ½50 vloop2
2/2 6/6 2/2
4/4
mg + FN = m v2 / r
50(10) + FN = 50 17.32 / 10
FN = 1000 N
6/6
= mghloop + ½mvloop2
= 500(2R) + ½50 14R
= 1350R
A 0.500 kg block rests on a frictionless table and is attached to a horizontal spring (k = 30 N/m) that is
uncompressed. A 0.100 kg wad of putty is launched horizontally on the frictionless table by another
identical spring which is compressed by 10 cm. The wad impacts the blocks and sticks. How far does the
putty-block system compress the original spring?
Workspring = ΔK
po
=
pf
Workspring =
ΔK
5/5
4/4
2
7/7
2
½kx =½mv
½30(.1)2 = ½(.1)vwad2
vwad = 1.73 m/s
7/7
5/5
5/5
2
½ k x = ½(mblock + mwad) vf2
½30x2 = ½(mblock + mwad) vf2
x = 4.08 cm
mvblock + mvwad = (mblock + mwad) vf
0 + .1 (1.73) = (0.5 + 0.1) vf
vf = 0.289 m/s
The collision between a sledge hammer and a nail for a rail tie can be considered to be approximately
elastic. Calculate the kinetic energy acquired by a 500-g nail when it is struck by a 8 kg hammer moving
with an initial speed of 5.0 m/s.
8/8
8/8
mh vh + mn vn = mh vhf + mn vnf
8 (5) + ½ (0) = 8vhf + ½vnf
80 - 16vhf = vnf
4/4
4/4
2
½mn vn2
2
4/4
= ½ mh vhf2 + ½mn
= ½ 8 vhf2 + ½ ½
=
vnf2
= (80 - 16 vhf)2
½mh vh +
½8 52 + ½ ½ 0
400 - 16 vhf2
400 - 16 vhf2
Vhf = 4.41 m/s; Vnf = 9.44 m/s
5/5
2
vnf
vnf2
Ko = 0
Kf = ½ m v2
Kf = ½ ½ 9.442
Kf = 22.3 J