PHYSICS 111 HOMEWORK
SOLUTION #7
March 10, 2013
0.1
A bead slides without friction around a looptheloop (see figure below).
The bead is released from rest at a height h = 3.30R.
• a) What is its speed at point A (Use the following as necessary:
the acceleration due to gravity g, and R.)
• b)How large is the normal force on the bead at point A if its
mass is 5.20 grams?
a)
In the absence of friction, the energy of the bead is conserved so that on each
point of the track we have:
mv 2
+ mgH = Constant
2
0 + mgh =
1
mv 2
2 A
vA
2
1
mv 2 + mg × 2R
2 A
= mg(3.30R − 2R)
p
=
2gR(3.30 − 2)
p
=
2.6gR
0.2.
b)
On point A, the centripetal force is the sum of gravity force and the normal
force:
N + mg
N
mv 2
gR
= m × 2.6
R
R
= mg(2.6 − 1)
=
=
5.20 × 9.81 × 1.6 × 10−3
=
0.018N
0.2
Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass m1 =
6.60 kg is released from rest at a height h = 3.40 m above the table.
• a)Using the isolated system model, determine the speed of the
object of mass m2 = 3.00 kg just as the 6.60kg object hits the
table.
• b)Find the maximum height above the table to which the 3.00kg
object rises.
a)
In the isolated system assumption, the total energy of the whole system is
conserved. Let’s take the table as our reference for heights. At rest, the total
energy is just the potential energy due to gravity of mass m1 that is m1 gh.
2
2
Once mass m1 hits the table this energy becomes : m12v + m22v + m2 gh
3
N.B. The two masses are moving with the same speed due to the strained
string.
Therefore,
m1 gh =
v
1
1
m1 v 2 + m2 v 2 + m2 gh
2
2
s
2gh(m1 − m2 )
m1 + m2
=
r
=
=
2 × 9.81 × 3.40 × (6.60 − 3.00)
6.60 + 3.00
5.00m/s
b)
Mass m1 still on the table, Mass m2 will keep ascending till its speed becomes
zero where it reaches a height H. Energy is again conserved between the moments m2 was at height h and height H:
m2 gH
H
4
= m2 gh +
= h+
m2 v 2
2
v2
2g
=
3.40 +
=
4.67m
52
2 × 9.81
0.3.
0.3
The coefficient of friction between the block of mass m1 = 3.00 kg
and the surface in the figure below is µk = 0.395. The system starts
from rest. What is the speed of the ball of mass m2 = 5.00 kg when
it has fallen a distance h = 1.70 m?
Due to friction, energy is not conserved. We can follow the same steps we used
in problem 3 Homework # 5 to dervive the acceleration of the two masses:
a=g
(m2 − µk m1 )
5.00 × −0.395 × 3.00
= 9.81 ×
= 4.68m/s2
m1 + m2
5.00 + 3.00
We can use the time-independent equation :
vf2 − vi2
vf
=
2ah
√
2ah
=
√
=
2 × 4.68 × 1.70
=
3.99m/s
5
0.4
A 5.20kg block is set into motion up an inclined plane with an initial
speed of vi = 8.40 m/s (see figure below). The block comes to rest
after traveling d = 3.00 m along the plane, which is inclined at an
angle of θ = 30.0◦ to the horizontal.
• a) For this motion, determine the change in the block’s kinetic
energy.
• b)For this motion, determine the change in potential energy of
the blockEarth system.
• c) Determine the friction force exerted on the block (assumed
to be constant).
• d) What is the coefficient of kinetic friction?
a)
+y
~
N
θ
f~k
+x
θ
m~g
6
0.4.
The change in kinetic energy is just:
1
1
mv 2 − mv 2
2 f 2 i
=
=
5.20 × 8.402
2
−183.46J
0−
b)
The change in potential energy is just:
mghf − mghi
= mgd sin θ
=
=
5.20 × 9.81 × 3 × sin 30◦
76.5J
c)
The change in the total energy is the work done by friction forces:
Wf riction = f~k · d~ =
∆Etotal
−fk × d =
fk
∆Etotal
∆Etotal
= −
d
−183.46 + 76.5
= −
3
= 35.7N
d)
fk = µk N
µk
=
µk mg cos θ
fk
=
mg cos θ
35.7
=
5.20 × 9.81 × cos 30
= 0.808
7
0.5
A child’s pogo stick as shown below stores energy in a spring with
a force constant of 2.55 × 10−4 N/m. At position A (xA =-130m) ,
the spring compression is a maximum and the child is momentarily
at rest. At position B , the spring is relaxed and the child is moving
upward. At position C, the child is again momentarily at rest at the
top of the jump.
The combined mass of the child and the pogo stick is 27.5 kg. Although the boy must lean forward to remain balanced, the angle is
small, so let’s assume the pogo stick is vertical. Also assume the boy
does not bend his legs during the motion.
• a)Calculate the total energy of the child-stick-Earth system,
taking both gravitational and elastic potential energies as zero
for x = 0.
• b)Determine xC
• c) Determine the value of x for which the kinetic energy of the
system is a maximum.
• d) Calculate the child’s maximum upward speed.
The Total Energy will be conserved
throughout the whole path
8
0.5.
a)
The system is at rest at point A (no kinetic energy), the total energy is just
the potential energy:
Etotal
=
=
=
1 2
kx + mgxA
2 A
1
× 25500(−0.130)2 + 27.5 × 9.81 × (−0.130)
2
180.40J
b)
C is another point of rest and the total energy is just the potential energy.
Etotal = 180.40
=
=
1 2
kx + mgxC
2 C
25500 2
xc + 27.5 × 9.81xC
2
Solving this equation leads to two values of xC : x =-0.130m which is the lower
extermum (Point A), and xC =0.109m.
c)
Another way to see Energy conservation law is to understand that that what’s
gained as a potential energy is lost as a kinetic energy and vice versa. Kinetic
energy is at its maximum when potential energy is at its minimum.
1 2
kx + mgx
2
The minium is to be found by the condition :
Epot =
dEpot (x)
dx
kx + mg
=
0
=
0
x =
−
=
=
mg
k
27.5 × 9.81
−
25500
−10.6mm
The position x = −10.6mm found above is the point of maximum kinetic
energy. Energy conservation will give:
Etotal = 180.40J
Ekin
=
Ekin + Epot
=
Ekin + mgx
=
180.40 − 27.5 × 9.81 × (−0.0106)
=
183.25J
9
and the maximum upward speed is :
r
v
=
=
=
2Ekin
m
r
2 × 183.25
27.5
3.65m/s
0.6
A 10.0kg block is released from rest at point A in the figure below.
The track is frictionless except for the portion between points B and
C, which has a length of 6.00 m. The block travels down the track, hits
a spring of force constant 2 200 N/m, and compresses the spring 0.200
m from its equilibrium position before coming to rest momentarily.
Determine the coefficient of kinetic friction between the block and the
rough surface between points B and C.
We choose the horizontal surface as the 0 potential energy.
2
Between A and B, energy is conserved and it amounts to : mgh = 12 mvB
Between B and C, energy is not conserved due to friction. The change in energy
2
2
is the work done by friction : 12 mvC
− 12 mvB
= f~k · d~ = −µk N d = −µk mgd
Remember N = mg from 2nd Law.
Between C and the final position, energy is conserved and it amounts to
1
1
2
2
2 mvC = 2 k(∆x) ; ∆x being the spring compression distance.
10
0.7.
All in all,
−µk mgd =
=
µk
=
=
=
1
1
2
2
mvC
− mvB
2
2
1
k(∆x)2 − mgh
2
h k(∆x)2
−
d
2mgd
3
2200 × 0.22
−
6 2 × 10 × 9.81 × 6
0.425
0.7
A block of mass m1 = 18.0 kg is connected to a block of mass m2 =
32.0 kg by a massless string that passes over a light, frictionless pulley.
The 32.0-kg block is connected to a spring that has negligible mass
and a force constant of k = 220 N/m as shown in the figure below.
The spring is unstretched when the system is as shown in the figure,
and the incline is frictionless. The 18.0-kg block is pulled a distance h
= 24.0 cm down the incline of angle θ = 40.0◦ and released from rest.
Find the speed of each block when the spring is again unstretched.
Let’s use the isolated system model consisting of the two blocks. Energy conservation requires that the change in potential energy for the system m1 + m2
is of same magnitude as the change in kinetic energy but with oposite sign.
∆Kin = −∆P ot
11
The two blocks are connected and move with the same speed v, we should have:
1
1
m1 v 2 + m2 v 2
2
2
=
(m1 + m2 )v 2
=
v
=
1
−(m1 g∆H1 + m2 g∆H2 + 0 − k(∆H2 )2 )
2
1
−(m1 gh sin θ − m2 gh − kh2 )
2
s
2gh(m2 − m1 sin θ) + kh2
m1 + m2
r
=
=
2 × 9.81 × 0.24(32 − 18 sin 40) + 220 × (0.24)2
32 + 18
1.48m/s
0.8
An electric scooter has a battery capable of supplying 105 Wh of
energy. If friction forces and other losses account for 60.0% of the
energy usage, what altitude change can a rider achieve when driving
in hilly terrain, if the rider and scooter have a combined weight of
930 N?
Let’s convert the supplied energy to joules.
105Wh = 105 × 3600W.seconds = 378000J
Only 40% will be available for the rider:
40% × 378000 = 151200J
This amount of energy is then used to reach a height h, the gain in potential energy is provided by the 151200 Joules supplied by the scooter battery.
Therefore,
mgh =
151200J
151200
h =
930
= 163m
12
0.9.
0.9
A 0.18-kg stone is held 1.4 m above the top edge of a water well and
then dropped into it. The well has a depth of 4.7 m.
• a)Relative to the configuration with the stone at the top edge
of the well, what is the gravitational potential energy of the
stoneEarth system before the stone is released.
• b)Relative to the configuration with the stone at the top edge
of the well, what is the gravitational potential energy of the
stoneEarth when it reaches the bottom of the well?
• c)What is the change in gravitational potential energy of the
system from release to reaching the bottom of the well?
a)
Relative to the chosen configuration for potential energy, the stone is at height
h1 = 1.4m before its release and the gravitational potential energy is:
Ep1
= mgh1
=
0.18 × 9.81 × 1.4
=
2.47J
b)
Relative to the chosen configuration for potential energy, the stone is at height
h2 = −4.7m at the bottom of the well and the gravitational potential energy
is:
Ep2
= mgh2
= −0.18 × 9.81 × 4.7
= −8.29J
13
c)
From its release to reaching the bottom, the gravitational potential energy
change is:
∆Ep
14
=
Ep2 − Ep1
=
−8.29 − 2.47
=
−10.8J