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Energy Problems
Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)
0J
60J
80J
600J
2400J
Since the Force is perpendicular
to the displacement, there is no
work done by this force.
Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)
27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K
1
 m  v 2f  vi2 
2
  m 2  m 2 
1
  4kg    6    3  
 s   s  
2


 54 J
Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)
-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height
Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?
a)
b)
c)
d)
e)
πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.
Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the book?
a)
b)
c)
d)
e)
-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement is upward
W  mgh
m

   2kg   9.8 2  1.5m 
s 

 30 J
Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?
a)
b)
c)
d)
e)
4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
The work done by gravity is
equal to the change in Kinetic
Energy.
W  K
1
mgh  m  v 2f  vi2 
2
1
gh  v 2f
2
v  2 gh
m

 2  9.8 2  sin  30  6m 
s 

m
 7.7
s
Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)
4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
Ki  U i  W f  K f  U f
1 2
mv f  0
2
1
mg  L sin       mg cos    L  v 2f
2
0  mgh  Ff L 
v f  2 gL  sin     cos   
This is a conservation of
Mechanical Energy, including
the negative work done by
friction.
m

 2  9.8 2   6m   sin  60   0.3cos  60  
s 

m
 9.2
s
Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?
a)
b)
c)
d)
e)
1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.
K  v2  v 
1
2
Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?
a)
b)
c)
d)
e)
50W
100W
200W
400W
Cannot be determined with given information
Use the power equation
P  Fv
 m
  200 N   2 
 s
 400W
Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height
pushing
lifting
10.0m
300
Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
lifting
10.0m
300
W  F d
 mgh
m

  2.0kg   9.8 2  10.0m 
s 

 196J
 200J
Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height
pushing
10.0m
300
The distance travelled up the ramp
d
10.0m
 20m
sin  30 
Applied Force
Fapplied  mg sin  
Work
W  Fapplied  d
W  mg sin   d
m

W   2.0kg   9.8 2  sin  30  20m 
s 

W  200J
Question
A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg
mass sitting still. Given that the collision is perfectly elastic, determine
the final velocities of each of the masses.
Solutions Question
A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg
mass sitting still. Given that the collision is perfectly elastic, determine
the final velocities of each of the masses.
Conservation of Momentum
pi  p f
m1v1i  m2v2i  m1v1 f  m2v2 f
 4.0kg   2.0
m
  0   4.0kg  v1 f  10kg  v2 f
s


kg  m
8.0
  4.0kg  v1 f  10kg  v2 f
s
Conservation of Energy
i  K2f
Oh No, we K
have
1 unknowns.
1
1
1
2
2If only
2
2
mthere
v

m
v

m
v

m
v
1 1i
2
2
i
1
1
f
2
2f
was
2
2 another
2
2
equation. 2
m
 4.0kg   2.0   0   4.0kg  v 21 f  10kg  v 2 2 f
s

16.0 J   4.0kg  v 21 f  10kg  v 2 2 f
Question
A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg
mass sitting still. Given that the collision is perfectly elastic, determine
the final velocities of each of the masses.
8.0
kg  m
  4.0kg  v1 f  10kg  v2 f
s
kg  m
8.0
  4.0kg  v1 f  10kg  v2 f
s
kg  m
8.0
 10kg  v2 f
s
v1 f 
4.0kg
16.0J   4.0kg  v21 f  10kg  v22 f
kg  m

8.0
 10kg  v2 f

s
16.0 J   4.0kg  
4.0kg


0  7v22 f  8v2 f
0  v2 f 7v2 f  8
m
v2 f  1.1
s
2


2
  10kg  v 2 f


kg  m
m

 10kg  1.14 
s
s

v1 f 
4.0kg
 0.82
8.0
Example 0
A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.
a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?
Solution
A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.
a) How much work is done by gravity?
Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp
Fg
Wgravity  F  d
  mg sin    d
m

  35kg   9.8 2  sin  40  8m 
s 

 1760 J
Solution
A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.
b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.
Solution
A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.
c) How much work is done by friction?
Recall Work = force x distance.
W friction   Ff  d
   uk mg cos    d
m

   0.3 35kg   9.8 2  cos  40  8m 
s 

 630 J
Solution
A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.
d) What is the total work done?
Total Work = sum of all works
Wtotal  Wgravity  WNormal  W friction
 1760 J  0 J  630 J
 1130 J
Example 2
A 2.00kg block is pushed against a spring with negligible mass and a force
constant k=400 N/m, compressing it 0.220 m. when the block is released, it
moves along a frictionless horizontal surface and then up a frictionless incline
with a slope of 37.00.
a) What is the speed of the block as it slides along the horizontal surface after
having left the spring?
b) How far does the block travel up the incline before starting to slide back
down?
Solution to Example 2
A 2.00kg block is pushed against a spring with negligible mass and a force constant
k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a
frictionless horizontal surface and then up a frictionless incline with a slope of 37.00.
a) What is the speed of the block as it slides along the horizontal surface after having left
the spring?
b) How far does the block travel up the incline before starting to slide back down?
Let’s find the Spring Potential
Energy first.
1 2
kx
2
1
N
2
  400   0.220m 
2
m
 9.68 J
US 
When released the Spring
Potential Energy is transferred
into kinetic energy of the Block,
where we can solve for v
1 2
mv
2
1
9.68 J   2.00kg  v 2
2
EK 
v
2  9.68 J 
2.00kg
 3.11
m
s
Solution to Example 1
A 2.00kg block is pushed against a spring with negligible mass and a force constant
k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a
frictionless horizontal surface and then up a frictionless incline with a slope of 37.00.
a) What is the speed of the block as it slides along the horizontal surface after having left
the spring?
b) How far does the block travel up the incline before starting to slide back down?
We can use conservation of
Spring Potential and
Gravitational Potential Energy
UG  U s
1 2
kx
2
mgh  9.68 J
mgh 
9.68 J
m
 2.00kg   9.8 2 
s 

 0.494m
h
Using Trig, we will translate
the vertical distance to a
ramp distance
0.494m
d
0.494m
d
sin  37.0 
sin  37.0  
 0.821m
Question 2
The planet Saturn is moving in the negative x-direction at its orbital
speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is
5.69x1026 kg. A 2150 kg spacecraft approaches Saturn, moving initially
in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a
conservative force) causes the spacecraft to swing around it and head of
into the opposite direction.
Determine the speed of the spacecraft after it is far enough away to be
free of Saturn’s gravitational pull.
Solution to Question 2
The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the
Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A 2150 kg spacecraft approaches
Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn
(a conservative force) causes the spacecraft to swing around it and head of into the opposite
direction.
Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s
gravitational pull.
Here the “collision” is not an impact but rather a
gravitational interaction. So we can treat it as a
one-dimensional elastic collision. Therefore the
relative velocities before and after the collision
have the same magnitude but opposite sign.
v pi  vsi    v pf  vsf

vsf  v pi  vsi  v pf
km
km 
km 
 10.4
  9.6

s
s 
s 
km
 29.6
s
 9.6
The reason the spacecraft
acquires so much additional
speed is that Saturn is moving in
its orbit. It does slow down, but
its mass is so much greater than
the spacecraft, it is not
noticeable.
Question 3
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?
Solution to Question 3
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?
Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W  ET
  K f  U f    Ki  U i 
Wp  K f  Ki  U f  U i
1 2 1 2
mv f  mvi  mgh  0
2
2
2
2

1
m 
m 
m

  80.0kg  1.50    5.00    80kg   9.8 2   5.2m 
2
s 
s  
s 


 3166.8 J

 3.17 103 J
Question 4
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
mv1i  Mv2i  mV  MV
 0.001kg  v1  0   3.001.kg V
v1  3001V
Stage 2 (conservation of Energy)
Ki  U f
1
 m  M V 2   m  M  gh
2
m

V 2  2  9.8 2   0.02m 
s 

m
V  0.626
s
Therefore:
m

v1  3001 0.626 
s

m
 1879
s
Question 5
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?
Solution to Question 5
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?
From Conservation of Energy
K1  U1  K 2  U 2
1 2
mv2  0
2
v2  2 gR
0  mgR 
m

 2  9.8 2   3.00m 
s 

m
 7.67
s
Solution to Question 5
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?
The free body diagram of the normal
is through-out his journey is:
F
y
 mac
v22
FN  FG  m
R
 2 gR 
FN  mg  m 

 R 
 mg  2mg
 3mg
v2  2 gR
Solution to Question 5
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?
The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.
The free body diagram of the normal
is through-out his journey is:
K1  U1  WF  K 2  U 2
WF  K 2  U 2  K1  U1

1 2
mv2  0  0  mgh
2
2
1
m
m


  25.0kg   6.00    25.0kg   9.80 2   3.00m 
2
s
s 


 285 J
Question 6
A 2000 kg elevator with broken cables is falling at 25 m/s when it first
contacts a cushioning spring at the bottom of the shaft. The spring is
supposed to stop the elevator, compressing 3.00 m as it does. During
the motion a safety clamp applies a constant 17,000 N frictional force to
the elevator.
a) Determine the force constant of the spring.
Solution to Question 6
A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning
spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing
3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional
force to the elevator.
a) Determine the force constant of the spring.
The elevator’s initial kinetic energy is:
2
1
1
 m
K1  mv12   2000kg   25   625, 000 J
2
2
 s
If we use Point 1 to be the origin, then we
have Us=0 and Ugrav=0, and so U1=0
At Point 2, there is both gravitational and
elastic potential energy, but no kinetic
energy
1
U 2  mgy2  ky22
2
Putting the conservation of energy to work:
K1+U1+Wother=K2+U2
Solution to Question 6
A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning
spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing
3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional
force to the elevator.
a) Determine the force constant of the spring.
K1  625,000 J
If we use Point 1 to be the origin, then we
have Us=0 and Ugrav=0, and so U1=0
1
U 2  mgy2  ky22
2
K1  U1  Wother  K 2  U 2
1
625, 000 J  0 J  Wother  0 J  mgy2  ky22
2

m


2  625, 000 J  17, 000 N  3.00m    2000kg   9.80 2   3.00m  
s 


k 
2
 3.00m 
 1.41105
N
m
Example 7
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
1.0 kg
Solution 7
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  uk FN
 uk mg
m

  0.10 1.0kg   9.8 2 
s 

 0.98N
Now apply the work Energy
Theorem and solve for d
Ff
W  KE
1
1
Ff d  mv 2f  mvi2
2
2
mg
2
1
m
1
m
 0.98 N  d  1.0kg   0   1.0kg   5.0 
2
s 
 s 2

12.5 J
d
0.98 N
 13m
2
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
A
a)
b)
c)
d)
e)
Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)
H
uk
B
x
C
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H
Etotal  UG  K  mgH  0  mgH
A
H
U
1
G
2
 K 1  Etotal
2
1  1 2
mg  H   mv  mgH
2  2
1 2 1
mv  mgH
2
2
v  gH
uk
B
x
C
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.
Etotal  UG  K  mgH  0  mgH
A
U GB  K B  Etotal
1 2
0  mv  mgH
2
v 2  2 gH
H
v  2 gH
uk
B
x
C
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C
A
H
W  K
1 2 1 2
W  mv f  mvi
2
2
1 2
 mvi
2
1 2
Fd  mvi
2
1 2
mguk x  mvi
2
v  2 gH
vi2
uk 
2 gx
H

x
uk
B
x
C
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.
K B  U B  W f  KC  U C
mgH  0  Ff L  0  mgh
A
H h
uk 
L cos  
mgH  uk mg cos   L  mgh
H
H  uk cos   L  h
L
B
H h

x
uk
x
C
Example 8
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)
A
H
uk
U A  K A  Wf  U B  KB
1 2
U A  K A  W f  mvB
2
2
1 1

mgH  0  W f  m 
2 gH 
2 2

mgH
mgH  W f 
4
mgH
3
Wf 
 mgH   mgH
4
4
B
x
Example 9
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
It’s too difficult to
use Centripetal
Acceleration
UI  K f U f
1 2
mv  mgL sin  30 
2
1
1
mgL  mv 2
2
2
gL  v 2
mgL 
v  gL
30
L
h  L sin  30
Example 10
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
a) Assuming no damping, what is the speed of the block,
i) as it passes through the equilibrium point?
ii) when it has moved 5.0 cm from its release point?
b) Find the time after release that the block first passed through the position
x = 3.0 cm (on the opposite side of its equilibrium point).
0.16m
Example 10 Solution
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
a) Assuming no damping, what is the speed of the block,
i) as it passes through the equilibrium point?
We can solve this in
two different ways
1) Using Energies
1 2 1 2
kA  mv
2
2
0.16m
1 2 1 2
kA  mv
2
2
k
v
A
m
N
m  0.160m 

3.00kg
36.0
 0.554
m
s
Example 10 Solution
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
a) Assuming no damping, what is the speed of the block,
i) as it passes through the equilibrium point?
We can solve this in
two different ways
2) Using Velocity Function
 k 
k
v  A
sin 
t 
m
m


0.16m
¼ of T
T  2
We need only the time when the mass
is at the equilibrium point (ie x=0). We
can get that by setting the position
function = 0 and solving for time or by
using one quarter of the period.
m
k
v    0.16 
3
36
 0.16
1
2
T
4
4
t

1
4 3
36
3
m
 0.554
s
 36   1  
36
sin 
 3  4 3  
3



Example 10
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
a) Assuming no damping, what is the speed of the block,
ii) when it has moved 5.0 cm from its release point?
We can solve this in
two different ways
1) Using Energies
1 2 1 2 1 2
kA  mv  kx
2
2
2
5cm
1 2 1 2 1 2
kx  mv  kA
2
2
2
k 2
v
A  x2 

m
x is the position that the spring is stretched
from the equilibrium at that at that velocity.
This distance is 16cm – 5 cm= 11 cm.
N
m

3.00kg
36.0
 0.40
m
s
16 cm
 0.160m    0.110m 
2
2

Example 10
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
a) Assuming no damping, what is the speed of the block,
ii) when it has moved 5.0 cm from its release point?
We can solve this in
two different ways
2) Using Velocity Function
5cm
 36 
0.11  0.16 cos 
t 
 k 
k
3


16 cm
v  A
sin 
t 
m
 36 
 m 
cos 
t   0.6875
3
 v   0.16 36 sin  36 0.2346 
 

 

3
3
Setting the position functions to 0.11 m and
36

t  cos 1  0.6875 
solving for time
3
m
 k 
x  A cos 
t 
m


36
t  0.812755
3
t  0.2346
 0.40
s
Example 10
A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to
a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away
from the equilibrium point and then released.
b) Find the time after release that the block first passed through the position
x = 3.0 cm (on the opposite side of its equilibrium point).
This is a job for the
position function.
 k 
x  A cos 
t 
m


3 cm
 k 
x  A cos 
t 
m


 36 
0.030m  0.160m cos 
t 
3


cos


12t  0.188
12t  cos 1  0.188
t  0.508s
16 cm
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D
H
E
θ
B
C
F
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
A
K A  U A  KC  U c
D
H
E
The centripetal
acceleration is v2/r
θ
B
0  mgH 
C
F
1 2
mvC  mgr
2
vC2  2 g  H  r 
vC2 2 g  H  r 

r
r
2g  H  r 
aC 
r
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
H
E
θ
B
0  mgH 
C
1 2
mv  mg  r  r cos 180    
2
F
1 2
mv  mg  r  r cos   
2
1
mg H  r 1  cos     mv 2
2
0  mgH 
The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).


v  2 g  H  r 1  cos   
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?
D
H
E
θ
B
FC  FG  FN
C
When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.
F
v2
m  mg  0
r
vcut off 
rmg
m
 gr
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
K A  U A  KD  U D
D
H
E
C
B
1 2
0  mgH  mv  mg  2r 
2
F
mgH 
Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.

1
m  gr   mg  2r 
2
5
mgr
2
H
5
r
2
Example 14
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
K A  U A  KB  U B
A
1
0  mg (6r )  mvB 2  0
2
D
H
E
C
B
F
Ff x  6mgr
Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.
k mgx  6mgr
6r
6r
x

 12r
k 0.5