Transcript Examples and Hints in Chapter 7

Examples and Hints in
Chapter 7
Wild Monkey III
Tarzan (m =100 kg) grabs a
vine to swing to cross a
chasm. He starts a from
a cliff face that is 10 m
taller than the opposite
side and his vine is 8 m
long. The vine will break if
its tension exceeds 1200
N. Can Tarzan make the
swing without the vine
breaking?
8m
10 m
Draw a Free Body diagram
Centripetal
force,
At bottom of rope: 2
Mv /R
Tension, T
Weight of Tarzan=100*9.8 =980 N
mv 2
 T  mg
R
mv 2
100* v 2
T
 mg 
 980
R
8
But what is v 2 ?
Use the Work Energy Thm:
DU+DK-Wother=0

There are no non-conservative forces so
DU+DK=0
Or
K1+U1=K2+U2
Initially, Tarzan at rest (K1=0) but 10 m above
final position (U1=mgy=980*10=9800J)
Finally, Tarzan at lower potential (U2=0) but in full
“swing” (K2= ½ mv2=0.5*100*v2)
So 9800 = 0.5*100*v2
196=v2
So
mv 2
 T  mg
R
mv 2
100* v 2
T
 mg 
 980
R
8
But what is v 2 ?
v 2  196
100*196
T
 980  3430
8
Rope Snaps!
One final question might be at what
height does the rope snap?
mv 2
 T  mg
R
mv 2
100* v 2
T
 mg 
 980
R
8
If T=1200 then
100* v 2
1200-980=220=
8
220*8
v2 
 17.6
100
U1  9800  U 3  K 3
1
K 3  *100*17.6  880
2
U 3  9800  880  8920
8920  980* h
h  8920 / 980  9.10 m
Bad Coaster
A 1000 kg rollercoaster
starts with a velocity
of 10 m/s from the
top of a track that is
65 m high.
If frictional forces do 400 kJ of work on it
as it approaches the
top of the loop-deloop which has a
radius of 10 m, is the
coaster able to make
the loop?
V1= 10 m/s
65 m
R=10 m
Draw a free body diagram at the top
of the loop
n, normal
force
(because
upside down)
mv 2

  n  mg
R
mv 2
1000* v 2
n
 mg 
 9800
R
10
But what is v 2 ?
Weight of coaster, mg=1000*9.8=9800
Use the Work Energy Thm:
DU+DK-Wother=0
U1=mg*65=9800*65=637 kJ
 K1= ½ *1000*10^2=50 kJ
 U2= mg*2*R=9800*20=196 kJ
 K2= ½ *1000*v2
 U1+K1-Wfriction=K2+U2
 637 kJ+50 kJ-400 kJ=500v2 + 196 kJ
 91=500v2
 V2=0.182

So
1000* v 2
n
 9800
10
n  100*.182  9800
n  18.2  9800
n  9782 N so car falls
Problem 7.74
A 2.00 kg package is released
on a 53.10 incline, 4.00 m
from a long spring with a
force constant of 120 N/m
that is attached to the
bottom of an incline. The
coefficients of friction
between the package and
the incline is ms=0.4 and
mk=0.2. The mass of the
spring is negligible.
a)
What is the speed of the
package just before it
reaches the spring?
b)
What is the maximum
compression of the
spring?
4m
53.10
Wgravity=2*9.8*4*cos(37.90)
Wgravity=62.72J
4m
Mg cos(530)
5
37.90
4
53.10
f=mk*mg*cos(530)
f=0.2*2*9.8*.6
f=2.354 N
Wfriction=2.354*4=9.408 J
3
mg
Use Work Energy
Net Work=Wgravity-Wfriction
=62.72-9.408=53.3 J
DK=Net Work
½ mv2 – 0=53.3
V=7.3 m/s
Part b) Use Work Energy
-DU=Wnet
Wfriction=2.354*(4+d)=9.408+2.354*d
Wgravity=15.67*(4+d)=62.69+15.67*d
Wnet=53.3J +13.3*d
-DU=0-1/2 *120*d2=-60d2
So
0=60d2+13.3*d+53.3
d=1.06 m
Hints
7.26— Concentrate only on the y-coordinate
7.30— Calculate the work for each path and then sum
7.39— Part a) Each rope carries 1/3 of the weight Part b) Calculate the total
work done against gravity Part c) Calculate the total path of the rope
7.42—Part a) Use DK=-DU to calculate Part B) Use W =-DU to calculate
where W=F*d*cosine(angle between)
7.46—Part a) Stay on the track the weight must equal the centripetal force
Part b) Ignore the path, concentrate on the difference in heights Part c)
Use velocity found in part b to compute
7.54—Use the conservation of energy that K+U-Wfriction=E, the total
mechanical energy and this E is conserved i.e. K1+U1-Wfriction=U2 in this
case
7.66—Use K1+U1=K2+U2-Wfriction
7.62—The skier’s kinetic energy at the bottom can be found by from the
potential energy at the top minus the work done by friction. Part b) in the
horzintoal, the potential energy is zero so K1=K2-Wfriction-Wair Part c)
F*d=DK