Introduction to Physics II
PHYS120
Chapter 6
Energy
Chapter 6: Energy
6.1 Work
6.2 Kinetic Energy and the Work–Energy Theorem
6.3 Gravitational Potential Energy
6.4 Spring Potential Energy
2
Introduction
Although energy can be
transformed from one kind to
another,
the total amount of energy in the
Universe or in an isolated system
never changes.
Various forms in the Universe:
Mechanical
Chemical
Electromagnetic
Nuclear
Energy
Kinetic energy:
the energy associated with motion
Potential energy:
the energy associated with relative position
Much easier than using forces and
Newton’s three laws
These two very different
approaches are linked through the
concept of work
3
6.1 Work
Work is done only if an object is moved through some
displacement while a force is applied to it
4
6.1 Work
The work W done on an object by a constant force 𝐹Ԧ
during a linear displacement ∆𝑥Ԧ is given by:
Work is a scalar
𝑊 has no direction
𝐹Ԧ and ∆𝑥Ԧ make an acute angle
0 <  < 90  𝑐𝑜𝑠 𝜃 > 0  W > 0
𝐹Ԧ and ∆𝑥Ԧ make an obtuse angle
90 <  < 180  𝑐𝑜𝑠 𝜃 < 0  W < 0
𝐹Ԧ and ∆𝑥Ԧ are perpendicular
𝑊 = (𝐹𝑐𝑜𝑠 𝜃)∆𝑥
 = 90  𝑐𝑜𝑠 𝜃 = 0  W = 0
𝐹Ԧ and ∆𝑥Ԧ in the same direction
F: is the magnitude of the force
x: is the magnitude of the object’s displacement
: is the angle between the directions of 𝐹Ԧ and ∆𝑥Ԧ
 = 0  𝑐𝑜𝑠 𝜃 = 1  W = 𝐹∆𝑥
SI unit: joule (J) = newton  meter = kg.m2/s2
 = 180  𝑐𝑜𝑠 𝜃 = −1  W = −𝐹∆𝑥
𝐹Ԧ and ∆𝑥Ԧ in opposite directions
5
6.1 Work
Work Is Done by
Something, on
Something Else:
Work doesn’t happen
by itself. Work is done
by something in the
environment, on the
object of interest.
A person holds a briefcase.
No work is done because the
displacement is zero
The person walks horizontally
while holding the briefcase. No
work is done because cos θ is zero.
6
6.1 Work
When a student lifts a box,
the work he does on the
box is positive because the
force he exerts on the box
is upward, in the same
direction as the
displacement.
When a student lowers a
box, the work he does on
the box is negative
because the force he exerts
on the box is upward, in
opposite direction to the
displacement.
A person pushes a lawn mower with a
constant force. The component of the force
parallel to the displacement is the work done
7
6.1 Work
QUICK QUIZ
In the Figures (a)–(d), a block moves to the right in the positive x-direction through the displacement x while under
the influence of a force with the same magnitude F. Which of the following is the correct order of the amount of work
done by the force F, from most positive to most negative?
(A) d, c, a, b
(B) c, a, b, d
(C) c, a, d, b
8
6.1 Work
Example 1:
An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled
and salmon is 50.0 kg, and the Eskimo exerts a force of 1.20  102 N on the sled by pulling on the rope.
(A) How much work does he do on the sled if the rope is horizontal to the ground ( = 0 in the Figure) and he
pulls the sled 5.00 m?
(B) How much work does he do on the sled if  = 30.0 and he pulls the sled the same distance of 5.00 m?
(C) Find the work done on the sled by friction, and the net work, when the force is exerted at  = 0 .
(Suppose the coefficient of kinetic friction between the loaded sled an snow is 0.200)
(D) Repeat the calculation of (C) if the applied force is exerted at an angle of 30.0 with the horizontal.
Answer:
(A)
𝑊 = 𝐹∆𝑥 = 1.2 × 102 𝑁 5𝑚 = 6 × 102 𝐽
(B)
𝑊 = (𝐹𝑐𝑜𝑠 𝜃)∆𝑥 = 1.2 × 102 𝑁 (cos 30°) 5𝑚 = 5.2 × 102 𝐽
9
6.1 Work
Example 1:
Answer:
(C) Apply the particle in equilibrium model to the sled:
In the y direction: ෍ 𝐹𝑦 = 0
→ 𝑛 − 𝐹𝑔 = 0
→ 𝑛 = 𝐹𝑔 = 𝑚𝑔
The work done by friction:
𝑊𝑓𝑟𝑖𝑐 = 𝑓𝑘 𝑐𝑜𝑠 𝜃 ∆𝑥 = −𝑓𝑘 ∆𝑥 = −𝜇𝑘 𝑛∆𝑥 = −𝜇𝑘 𝑚𝑔∆𝑥
𝑊𝑓𝑟𝑖𝑐 = −(0.2)(50𝑘𝑔)(9.8𝑚/𝑠 2 )(5𝑚)
𝑊𝑓𝑟𝑖𝑐 = −4.9 × 102 𝐽
The net work:
𝑊𝑛𝑒𝑡 = 𝑊𝑎𝑝𝑝 + 𝑊𝑓𝑟𝑖𝑐 + 𝑊𝑛 + 𝑊𝑔
𝑊𝑛𝑒𝑡 = 6 × 102 𝐽 − 4.9 × 102 𝐽 + 0 + 0
𝑊𝑛𝑒𝑡 = 1.1 × 102 𝐽
10
6.1 Work
Example 1:
Answer:
(D) Apply the particle in equilibrium model to the sled:
In the y direction: ෍ 𝐹𝑦 = 0
→ 𝑛 − 𝐹𝑔 + 𝐹𝑎𝑝𝑝 sin 𝜃 = 0
→ 𝑛 = 𝐹𝑔 − 𝐹𝑎𝑝𝑝 sin 𝜃 = 𝑚𝑔 − 𝐹𝑎𝑝𝑝 sin 𝜃
The work done by friction:
𝑊𝑓𝑟𝑖𝑐 = 𝑓𝑘 𝑐𝑜𝑠 𝜃 ∆𝑥 = −𝑓𝑘 ∆𝑥 = −𝜇𝑘 𝑛∆𝑥 = −𝜇𝑘 (𝑚𝑔 − 𝐹𝑎𝑝𝑝 sin 𝜃)∆𝑥
𝑊𝑓𝑟𝑖𝑐 = −(0.2)(50𝑘𝑔 × 9.8𝑚/𝑠 2 − 1.2 × 102 𝑁 × cos 30°)(5𝑚)
𝑊𝑓𝑟𝑖𝑐 = −4.3 × 102 𝐽
The net work:
𝑊𝑛𝑒𝑡 = 𝑊𝑎𝑝𝑝 + 𝑊𝑓𝑟𝑖𝑐 + 𝑊𝑛 + 𝑊𝑔
𝑊𝑛𝑒𝑡 = 5.2 × 102 𝐽 − 4.3 × 102 𝐽 + 0 + 0
𝑊𝑛𝑒𝑡 = 90𝐽
The most important thing to notice here is
that exerting the applied force at different
angles can dramatically affect the work
done on the sled. Pulling at the optimal
angle (11.3 in this case) will result in the
most net work for the same applied force.
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6.2 Kinetic Energy and the Work–Energy Theorem
If the net work can be calculated for a given displacement, the change in the object’s speed is easy to evaluate.
Because the force is constant, we know from Newton’s second law that the
object moves with constant acceleration 𝑎.
Ԧ
If the object is displaced by x, the work done by 𝐹Ԧ net on the object is
𝑊𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡 ∆𝑥 = (𝑚𝑎)∆𝑥
When an object undergoes constant acceleration:
𝑣 2 − 𝑣02
2
2
𝑣 = 𝑣0 + 2𝑎∆𝑥 𝑜𝑟 𝑎∆𝑥 =
2
𝑣 2 − 𝑣02
𝑊𝑛𝑒𝑡 = 𝑚
2
→ 𝑊𝑛𝑒𝑡 =
1
1
2
𝑚𝑣 − 𝑚𝑣02
2
2
The kinetic energy KE of an object of mass m moving with a speed v is
1
𝐾𝐸 ≡ 𝑚𝑣 2
2
The net work done on an object is equal to the change in the object’s kinetic energy:
𝑊𝑛𝑒𝑡 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 = ∆𝐾𝐸
where the change in the kinetic energy is due entirely to the object’s change in speed.
(Work–Energy Theorem)
12
6.2 Kinetic Energy and the Work–Energy Theorem
Example 2:
The driver of a 1.00  103 kg car traveling on the interstate at 35.0 m/s slams on his brakes to avoid
hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the
brakes are applied, a constant friction force of 8.00  103 N acts on the car. Ignore air resistance.
(A) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle?
(B) If the distance between the vehicles is initially only 30.0 m, at what speed would the collision occur?
Answer:
(A) Apply the work–energy theorem to the car:
1
1
2
𝑊𝑛𝑒𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖2
2
2
1
⟹ −𝑓𝑘 ∆𝑥 = 0 − 𝑚𝑣𝑖2
2
1
⟹ −(8.00  103 N)∆𝑥 = 0 − (1.00  103 kg)(35.0 m/s)2
2
⟹ ∆𝑥 = 76.6𝑚
13
6.2 Kinetic Energy and the Work–Energy Theorem
Example 2:
Answer:
(B) Apply the work–energy theorem to the car:
1
1
2
𝑊𝑛𝑒𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖2
2
2
1
1
2
⟹ −𝑓𝑘 ∆𝑥 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖2
2
2
⟹ 𝑣𝑓2 = 𝑣𝑖2 −
⟹ 𝑣𝑓2 =
2
𝑓 ∆𝑥
𝑚 𝑘
35.0 m/s 2 −
2
(8.00  103 N)(30.0 m)
3
(1.00  10 kg)
⟹ 𝑣𝑓2 = 745𝑚2 /𝑠 2
⟹ 𝑣𝑓 = 27.3m/s
This calculation illustrates how important it is to remain alert on the highway, allowing for an adequate
stopping distance at all times. It takes about a second to react to the brake lights of the car in front of you. On
a high speed highway, your car may travel more than 30 meters before you can engage the brakes. Bumper-tobumper traffic at high speed, as often exists on the highways near big cities, is extremely unsafe.
14
6.2 Kinetic Energy and the Work–Energy Theorem
Conservative and Non-conservative Forces
A force is conservative if the work it does
moving an object between two points is the
same no matter what path is taken.
Gravity field is conservative, so the diver
regains as kinetic energy the work she did
against gravity in climbing the ladder. Taking
the frictionless slide gives the same result.
A non-conservative force is generally dissipative,
which means that it tends to randomly disperse the
energy of bodies on which it acts.
Friction is a non-conservative force, so a
book pushed along the three segments A–B,
B–C, and C–D requires three times the work
as pushing the book directly from A to D.
15
6.3 Gravitational Potential Energy
Potential energy is a property of a system, rather than of a single object, because it’s due to a
physical position in space relative to a center of force
A brick on a high shelf can do work:
- It can fall off the shelf
- accelerate downwards
- hit the floor
The brick is said to have potential energy associated
with it, because from its location on the shelf it can
potentially do work.
16
6.3 Gravitational Potential Energy
The difference of the potential energy function at any two points gives the
negative of the work done by that force between those two points:
𝑊𝑔 = 𝐹∆𝑦𝑐𝑜𝑠 𝜃 = 𝑚𝑔 𝑦𝑖 − 𝑦𝑓 cos 0° = −𝑚𝑔 𝑦𝑓 − 𝑦𝑖
This equation for gravitational work holds for any object, regardless of its
trajectory in space, because the gravitational force is conservative.
The gravitational potential energy of a system consisting of Earth and an object
of mass m near Earth’s surface is given by
𝑃𝐸 ≡ 𝑚𝑔𝑦
• g: acceleration of gravity
• y: vertical position of the mass relative to the surface of Earth
(or some other reference point).
17
6.3 Gravitational Potential Energy
The zero level:
 Is usually taken to correspond to Earth’s surface, but this is not strictly necessary.
 Once chosen, it must remain fixed for a given problem.
 The difference in potential energy (the important quantity) will be the same regardless of the choice of zero level.
the floor might be a
more convenient
reference level.
A natural zero level for
potential energy is the
surface of the desk
Choose the surface of
Earth as the zero level
of potential energy.
18
6.3 Gravitational Potential Energy
Example 3:
A 60.0-kg skier is at the top of a slope, as shown in the Figure. At the initial point A, she is 10.0 m vertically above point B.
(A) Setting the zero level for gravitational potential energy at B:
i. Find the gravitational potential energy of this system when the skier is at A and then at B.
ii. Find the change in potential energy of the skier–Earth system as the skier goes from point A to point B.
(B) Repeat this problem with the zero level at point A.
(C) Repeat again, with the zero level 2.00 m higher than point B.
Answer:
(A) The zero level at point B
(i) The potential energy at A:
𝑃𝐸𝑖 = 𝑚𝑔𝑦𝑖 =(60.0kg )(9.8m/𝑠 2 )(10.0 m)=5.88 × 103 𝐽
The potential energy at B:
𝑃𝐸𝑓 = 0
(ii) The change in potential energy:
𝑃𝐸𝑓 − 𝑃𝐸𝑖 = 0 − 5.88 × 103 𝐽 = −5.88 × 103 𝐽
19
6.3 Gravitational Potential Energy
Example 3:
Answer:
(B) The zero level at point A
(i) The potential energy at A:
𝑃𝐸𝑖 = 0
The potential energy at B:
𝑃𝐸𝑓 = 𝑚𝑔𝑦𝑖 =(60.0kg )(9.8m/𝑠 2 )( − 10.0 m)= − 5.88 × 103 𝐽
(ii) The change in potential energy:
𝑃𝐸𝑓 − 𝑃𝐸𝑖 = −5.88 × 103 𝐽 − 0 = −5.88 × 103 𝐽
20
6.3 Gravitational Potential Energy
Example 3:
Answer:
(C) The zero level at 2.00 m higher than point B
(i) The potential energy at A:
𝑃𝐸𝑖 = 𝑚𝑔𝑦𝑖 =(60.0kg )(9.8m/𝑠 2 )(8.0 m)=4.7 × 103 𝐽
The potential energy at B:
𝑃𝐸𝑓 = 𝑚𝑔𝑦𝑖 =(60.0kg )(9.8m/𝑠 2 )( − 2.0 m)= − 1.18 × 103 𝐽
(ii) The change in potential energy:
𝑃𝐸𝑓 − 𝑃𝐸𝑖 = −1.18 × 103 𝐽 − 4.7 × 103 𝐽 = −5.88 × 103 𝐽
21
6.3 Gravitational Potential Energy
In any isolated system of objects interacting only through conservative forces, the total
mechanical energy E = KE + PE, of the system, remains the same at all times.
Conservation of
mechanical energy
Tossing a rock off a cliff:
(ignoring the drag forces)
• As the rock falls, its speed increases,
so its kinetic energy increases.
• As the rock approaches the ground,
the potential energy of the
rock–Earth system decreases.
Whatever potential energy is lost as the rock
moves downward appears as kinetic energy
22
6.3 Gravitational Potential Energy
Example 4:
A diver of mass m drops from a board 10.0 m above the water’s surface, as in the Figure. Neglect air resistance.
(A) Use conservation of mechanical energy to find his speed 5.00 m above the water’s surface.
(B) Find his speed as he hits the water.
Answer:
(A)
Conservation of mechanical energy:
𝐸𝑖 = 𝐸𝑓
𝐾𝐸𝑖 + 𝑃𝐸𝑖 = 𝐾𝐸𝑓 + 𝑃𝐸𝑓
1
1
𝑚𝑣𝑖2 + 𝑚𝑔𝑦𝑖 = 𝑚𝑣𝑓2 + 𝑚𝑔𝑦𝑓
2
2
1
0 + 𝑔𝑦𝑖 = 𝑣𝑓2 + 𝑔𝑦𝑓
2
𝑣𝑓 =
2𝑔(𝑦𝑖 − 𝑦𝑓 )
𝑣𝑓 =
2(9.8𝑚/𝑠 2 )(10𝑚 − 5𝑚)
𝑣𝑓 = 9.9𝑚/𝑠
23
6.3 Gravitational Potential Energy
Example 4:
Answer:
(B) Conservation of mechanical energy:
𝐸𝑖 = 𝐸𝑓
𝐾𝐸𝑖 + 𝑃𝐸𝑖 = 𝐾𝐸𝑓 + 𝑃𝐸𝑓
1
1
2
𝑚𝑣𝑖 + 𝑚𝑔𝑦𝑖 = 𝑚𝑣𝑓2 + 𝑚𝑔𝑦𝑓
2
2
1 2
0 + 𝑔𝑦𝑖 = 𝑣𝑓 + 0
2
𝑣𝑓 =
2𝑔𝑦𝑖
𝑣𝑓 =
2(9.8𝑚/𝑠 2 )(10𝑚)
𝑣𝑓 = 14𝑚/𝑠
24
6.4 Spring Potential Energy
Hooke’s law:
𝐹𝑠 = −𝑘𝑥
The minus sign ensures that the
spring force is always directed back
towards the equilibrium point.
Fs: is often called a restoring force
k: is a constant of proportionality, the spring constant
x: displacement
Spring Potential Energy:
Is the negative of the work done by the spring. It can also be considered as the
stored energy arising from the work done to compress or stretch the spring.
1 2
𝑃𝐸𝑠 ≡ 𝑘𝑥
2
25
6.4 Spring Potential Energy
Example 5:
A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00  102 N/m, as in the Figure. The
surface the block rests upon is frictionless. If the block is pulled out to xi = 0.050 m and released:
(A) Find the speed of the block when it first reaches the equilibrium point.
(B) Find the speed when x = 0.025 m.
Answer:
(A) Conservation of mechanical energy:
(𝐾𝐸 + 𝑃𝐸𝑔 + 𝑃𝐸𝑠 )𝑖 = (𝐾𝐸 + 𝑃𝐸𝑔 + 𝑃𝐸𝑠 )𝑓
1
1 2 1
1 2
2
2
𝑚𝑣𝑖 + 𝑘𝑥𝑖 = 𝑚𝑣𝑓 + 𝑘𝑥𝑓
2
2
2
2
1 2 1
𝑘𝑥𝑖 = 𝑚𝑣𝑓2
2
2
𝑣𝑓 =
𝑘
𝑥 =
𝑚 𝑖
𝑣𝑓 = 0.447𝑚/𝑠
4.00  102 N/m
(0.050 m)
5.00 kg
26
6.4 Spring Potential Energy
Example 5:
Answer:
(B) Conservation of mechanical energy:
(𝐾𝐸 + 𝑃𝐸𝑔 + 𝑃𝐸𝑠 )𝑖 = (𝐾𝐸 + 𝑃𝐸𝑔 + 𝑃𝐸𝑠 )𝑓
1
1 2 1
1 2
2
2
𝑚𝑣𝑖 + 𝑘𝑥𝑖 = 𝑚𝑣𝑓 + 𝑘𝑥𝑓
2
2
2
2
2
𝑘𝑥
𝑘𝑥𝑖2
𝑓
= 𝑣𝑓2 +
𝑚
𝑚
Therefore,
𝑣𝑓 =
𝑘 2
(𝑥𝑖 − 𝑥𝑓2 ) =
𝑚
4.00  102 N/m
(0.050𝑚2 − 0.0252 )
5.00 kg
𝑣𝑓 = 0.387𝑚/𝑠
27