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Phys Int CC Ch 9 - Energy - Answers PDF

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CK-12 Physics Concepts - Intermediate
Answer Key
Chapter 9: Energy
9.1 Kinetic Energy
Practice
Questions
1.
2.
3.
Potential energy is present in objects that are ____________.
Kinetic energy is present in objects that are ___________.
What formula is given for kinetic energy?
Answers
1.
2.
3.
Stationary.
Moving.
KE = (1/2) mv^2
Review
Questions
1.
2.
A comet with a mass of 7.85 × 1011 kg is moving with a velocity of 25,000 m/s.
Calculate its kinetic energy.
A rifle can shoot a 4.00 g bullet at a speed of 998 m/s.
a.
Find the kinetic energy of the bullet.
b.
What work is done on the bullet if it starts from rest?
c.
If the work is done over a distance of 0.75 m, what is the average force on
the bullet?
d.
If the bullet comes to rest after penetrating 1.50 cm into a piece of metal,
what is the magnitude of the force bringing it to rest?
Answers
1
1
π‘š
1.
Using 𝐾𝐸 = 2 π‘šπ‘£ 2 ∢ 𝐾𝐸 = 2 (7.85 × 1011 π‘˜π‘”) (25,0002 𝑠 ) ∢ The comet’s kinetic
energy is 2.45 × 1020 J.
2.
1
π‘š
a. Using 𝐾𝐸 = 2 π‘šπ‘£ 2 ∢ (. 004π‘˜π‘”) (9982 𝑠 ) = 1992 J. (Remember to convert
4.00 g to kg.)
b. π‘Šπ‘›π‘’π‘‘ = βˆ†πΎπΈ ∢ 1992 J.
𝑀
1992𝐽
c. Using 𝑑 = 𝐹 ∢ .75π‘š = 2656 N.
1
d. Using
𝑀
𝑑
=𝐹∢
1992𝐽
0.015π‘š
= 132,800N.
9.2 Potential Energy
Practice
Questions
1.
2.
3.
What is the definition of energy?
Name two types of potential energy.
How is energy transferred from one object to another?
Answers
1.
2.
3.
Energy is the ability or capacity to move an object (to do work).
Chemical, Gravitational, Elastic, Electrical, and Nuclear.
Energy is transferred from one object to another by doing work.
Review
Questions
1. A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How
much potential energy does he have at the top?
2. A 50.0 kg shell was fired from a cannon at earth’s surface to a maximum height
of 400m.
a. What is the potential energy at maximum height?
b. It then fell to a height of 100. m. What was the loss of PE as it fell?
3. A person weighing 645 N climbs up a ladder to a height of 4.55 m.
a. What work does the person do?
b. What is the increase in gravitational potential energy?
c. Where does the energy come from to cause this increase in PE?
Answers
π‘š
1. Using 𝑃𝐸 = π‘šπ‘”β„Ž ∢ 90.0π‘˜π‘” ∗ 9.80 𝑠2 ∗ 9.47π‘š ∢ 8350 J.
2.
π‘š
a. Using 𝑃𝐸 = π‘šπ‘”β„Ž ∢ 50.0π‘˜π‘” ∗ 9.80 𝑠2 ∗ 400π‘š = 196,000 J.
π‘š
b. Using 𝑃𝐸 = π‘šπ‘”β„Ž ∢ 50.0π‘˜π‘” ∗ 9.80 𝑠2 ∗ 100π‘š = 49,000𝐽 Subtract initial PE
from final PE : 196,000 − 49,000 = 147000 J.
3.
a. Using 𝑃𝐸 = (π‘šπ‘”)β„Ž ∢ 645𝑁 ∗ 4.55π‘š = 2930 J.
b. Starts from 0: 2930 J.
2
c. The energy increase is the same as the work done against the
gravitational force. The person at the top of the ladder has more distance
to fall and thus more potential energy than he/she would at the bottom of
the ladder.
9.3 Conservation of Energy
Practice
Questions
1.
2.
3.
What happens when one ball is pulled up to one side and released and why?
What happens when three balls are pulled up to one side and released and why?
What happens when two balls are pulled out from each side and released and
why?
Answers
1.
2.
3.
The potential energy of the ball is converted into kinetic energy as it falls. When it
hits the four stationary balls, the kinetic energy is transferred through them to the
last ball which rises to the same height as the starting ball, converting the kinetic
energy back into potential energy. The process is then repeated in the opposite
direction.
When three balls are pulled up, the potential energy is again converted into
kinetic energy. This time, the kinetic energy of the leftmost three balls is
transferred into the rightmost three balls. The kinetic energy is converted back
into potential energy and the process is repeated in the opposite direction.
When two balls are pulled out from each side and released, they fall as before.
The kinetic energy from both directions impacts the stationary balls at the same
time. The central balls have no net force and so they remain stationary. The
outer balls bounce back as if they had hit a solid wall and bounced off. They lose
kinetic energy and gain potential energy as they rise from the impact point.
Review
Questions
1. A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00
m to the surface of the water:
a. what is the kinetic energy of the chunk of ice when its hits the water, and
b. what is its velocity?
2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a
hill and coasts up the hill.
a. Assuming the movement is frictionless, at what vertical height will the cart
come to rest?
3
3.
4.
5.
6.
b. Do you need to know the mass of the cart to solve this problem?
A circus performer swings down from a platform on a rope tied to the top of a tent
in a pendulum-like swing. The performer’s feet touch the ground 9.00 m below
where the rope is tied. How fast is the performer moving at the bottom of the arc?
A skier starts from rest at the top of a 45.0 m high hill, coasts down into a valley,
and continues up to the top of a 40.0 m high hill. Both hill heights are measured
from the valley floor. Assume the skier puts no effort into the motion (always
coasting) and there is no friction.
a. How fast will the skier be moving on the valley floor?
b. How fast will the skier be moving on the top of the 40.0 m hill?
A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m
high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is
the magnitude of the final velocity when it strikes the ground? Ignore air
resistance.
If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum
height will it reach? Neglect air resistance.
Answers
1.
π‘š
a. Using 𝑃𝐸 = π‘šπ‘”β„Ž ∢ 15π‘˜π‘” ∗ 9.8 𝑠2 ∗ 8π‘š ∢ 1180 J
2𝑑
2(8π‘š)
b. First solve for time to fall using: 𝑑 = √ 𝑔 ∢ √ π‘š = 1.3𝑠. Then solve for
9.8
𝑠2
π‘š
velocity using: 𝑣 = 𝑔𝑑 ∢ 9.8 𝑠2 ∗ 1.3𝑠 =12.5 m/s
2.
1
π‘š 2
1
a. Solve for KE using 𝐾𝐸 = 2 π‘šπ‘£ 2 ∢ 2 (85.0π‘˜π‘”) (9.00 𝑠 ) = 3442.5𝐽. Then
solve for height using 3442.5J as PE with:
𝑃𝐸
π‘šπ‘”
3442.5𝐽
= β„Ž ∢ 85π‘˜π‘”∗9.8 π‘š/𝑠2 = 4.13 m
b. No. To solve the problem, you equate the potential and kinetic energy
equations:
. Since mass is on both sides of the equation, it
cancels out. Thus, we can determine the maximum height based only on
the cart’s speed.
2𝑑
2(9π‘š)
3. First solve for time to fall using: 𝑑 = √ 𝑔 ∢ √ π‘š = 1.35𝑠. Then solve for velocity
9.8
𝑠2
π‘š
using: 𝑣 = 𝑔𝑑 ∢ 9.8 𝑠2 ∗ 1.35𝑠 =13.23 m/s The performer’s velocity is 13.2 m/s
at the bottom of the swing.
4.
2𝑑
2(45π‘š)
a. First solve for time to fall using: 𝑑 = √ 𝑔 ∢ √
π‘š
π‘š
9.8 2
𝑠
velocity using: 𝑣 = 𝑔𝑑 ∢ 9.8 𝑠2 ∗ 3.03𝑠 =29.7 m/s
4
= 3.03𝑠. Then solve for
b. Speed at the top of hill#2 is equal to speed at the bottom – acceleration
2𝑑
π‘š
2∗40π‘š
due to g over 40m. Acceleration over 40m = 𝑔 ∗ √ 𝑔 = 9.8 𝑠2 ∗ √
π‘š
π‘š
9.8 2
𝑠
=
28 𝑠 . Speed at the bottom of the hill = 29.7 m/s, subtract 28 m/s to get
speed at the top of hill#2: 1.7m/s
1
1
π‘š 2
5. Solve for final KE using 2 π‘šπ‘£ 2 + π‘šπ‘”β„Ž = 2 (2π‘˜π‘”) (20 𝑠 ) +
π‘š
2(792𝐽)
(2π‘˜π‘”) (9.8 2 ) (20.0π‘š) = 792𝐽. Then solve for final velocity using 𝑣𝑓 = √
𝑠
2.0π‘˜π‘”
=
√792 = 28.1 m/s.
𝑃𝐸
500
6. Using β„Ž =
=
= 25.51 m.
π‘šπ‘”
2∗9.8
9.4 Elastic and Inelastic Collisions
Practice
Questions
1. Explain what happened in the first demonstration on elastic collisions.
2. Explain what happened in the second demonstration on inelastic collisions.
3. Assuming the first carts started at the same speed in both demonstrations, explain
why the inelastic collision ended slower than the elastic collision.
Answers
1. The momentum from the first cart was completely transferred into the second when
the collision occurred and kinetic energy was unchanged.
2. The momentum from the first cart was shared between the two carts when the
collision occurred and kinetic energy decreased
3. The end product of the inelastic collision had more mass than the elastic collision.
Because momentum remained unchanged in both demonstrations but the mass of
the second increased, the velocity had to decrease or else the momentum would also
increase.
Review
Questions
1. A 4.00 kg metal cart is sitting at rest on a frictionless ice surface. Another metal
cart whose mass is 1.00 kg is fired at the cart and strikes it in a one-dimensional
elastic collision. If the original velocity of the second cart was 2.00 m/s, what are
the velocities of the two carts after the collision?
2. Identify the following collisions as most likely elastic or most likely inelastic.
a. A ball of modeling clay dropped on the floor.
5
b.
c.
d.
e.
A fender-bender automobile collision.
A golf ball landing on the green.
Two billiard balls colliding on a billiard table.
A collision between two ball bearings.
Answers
1. Solve for cart #1’s velocity using: π‘š1 = 4π‘˜π‘”, π‘š2 = 1π‘˜π‘”, 𝑣𝑖1 = 0
a. 𝑣𝑓1 =
(π‘š1 −π‘š2 )𝑣𝑖1 +2π‘š2 𝑣𝑖2
b. 𝑣𝑓2 =
2π‘š1 𝑣𝑖1 −(π‘š1 −π‘š2 )𝑣𝑖2
π‘š1 +π‘š2
π‘š1 +π‘š2
∢
∢
π‘š
𝑠
(4π‘˜π‘”−1π‘˜π‘”)0 +2(1π‘˜π‘”)(2)
4π‘˜π‘”+1π‘˜π‘”
π‘š
𝑠
π‘š
𝑠
(2∗4π‘˜π‘”∗0 )−(4π‘˜π‘”−1π‘˜π‘”)2
4π‘˜π‘”+1π‘˜π‘”
= .8
π‘š
𝑠
, 𝑣𝑖2 = 2.00
π‘š
𝑠
π‘š
𝑠
= −𝟏. 𝟐
π’Ž
𝒔
c. The first cart’s final velocity is 0.8 m/s, and the second cart’s final velocity
is -1.2 m/s.
2.
a.
b.
c.
d.
e.
Inelastic – change in kinetic energy because the clay will compress.
Inelastic – change in kinetic energy because both cars will compress.
Inelastic – change in kinetic energy because the grass will compress.
Elastic
Elastic
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