Texas Tech University, Department
of Physics
Physics 1308 Section 004
Part 1: Potential Energy and the Work
Energy Theorem

Potential energy is the energy associated with
the object’s POSITION in space

First kind: Gravitational Potential energy, U_{g}
= mgy, where y is the distance of the object from the zero reference point

This zero reference point is arbitrary, it is
not a fixed reference point.

Second kind: Elastic Potential energy, U_{s}
= ½ kx^{2}, where x is the distance from the equilibrium
or zero reference point which refers to the springs unstretched state.

Work –Energy theorem: W_{net} = DU
Problem 1: Potential energy and a
roller coaster. A 1000kg roller coaster car moves from point A to B then
C.
(a) What is the potential energy at B and C relative
to A?
Solution:
The reference point for this part is A, therefore our y = 0 is at point
A
Point B is a distance y_{b} = (30  10)m = 20 m below our reference
point
U_{b} = mgy_{b} = (1000 kg)(9.80 m/s^{2})(20 m)
= 196 kJ
Point C is a distance y_{c} = (30  25)m = 5 m below our reference
point
U_{c} = mgy_{c} = (1000 kg)(9.80 m/s^{2})(5 m)
= 49 kJ
(b) What is the change in potential energy
as it goes from B to C?
Solution:
DU
= U_{c} U_{b} = 49 kJ  (196 kJ) = 147 kJ
(c) What is the potential energy at B and
C relative to point C?
Solution:
The reference point for this part is C, therefore our y = 0 is at point
V
Point B is a distance y_{b} = (25  10)m = 15 m below our reference
point
U_{b} = mgy_{b} = (1000 kg)(9.80 m/s^{2})(15 m)
= 147 kJ
Point C is a distance y_{c} = 0
U_{c} = mgy_{c} = (1000 kg)(9.80 m/s^{2})(0) =
0
(d) What is the change in potential energy
as it goes from B to C?
Solution:
DU
= U_{c} U_{b} = 0 kJ  (147 kJ) = 147 kJ
Therefore, regardless of our choice of reference point, the change in potential
energy between 2 points will always be the same.
Problem 2: A spring with spring constant
k = 400 N/m, is stretched from it's equilibrium position by 1.5 m, find
the energy stored in the spring. If the same spring is compressed by an
amount of 1.5 m from the equilibrium, what would the energy be?
Solution:
U_{s} = ½ kx^{2 }=½(400 N/m)(1.5 m)^{2}
= 450 J regardless of whether it was compressed by 1.5 m or stretched by
1.5 m.
PART 2: Conservation
of Energy and Conservative Forces

A conservative force is a force that does work
on an object that is independent of the path taken. Therefore, it does
not matter how you get from point A to pint B, because the work done will
be the same regardless of the path taken.

Gravitational Force and Elastic forces are examples
of conservative forces.

Total Mechanical Energy E = K + U, just the sum
of the kinetic and the potential energy acting on a system.

If a system has only conservative forces acting
on it, then the conservation of mechanical energy states: the mechanical
energy of the system is constant, therefore DE
= 0 or E_{i} = E_{f }.

Simply put: the energy of the system anywhere
on its path is constant, the only thing that may change is the type of
energy involved.
Problem 1: A dart of mass m = 0.100
kg is pressed against a spring of a toy dart gun. The spring has a spring
constant k = 250 N/m, and it is compressed by a distance of x_{i}
= 0.06m. If the dart detaches from the spring once the spring reaches it's
equilibrium position, x_{f}=0, what is the speed of the dart?
Solution:
The initial energy of the system is purely elastic potential energy since
the spring is compressed and the mass is initially at rest. Therefore:
E_{i} = ½ kx^{2 }=½(250 N/m)(0.06 m)^{2}
= 0.45 J
The final energy of the system is purely kinetic since the mass is now
moving and the spring is now back at its equilibrium position. Therefore:
E_{f} = ½ mv^{2 }=½(0.100 kg)v^{2}
By conservation of energy: E_{i} = E_{f} therefore
½(0.100 kg)v^{2} = 0.45 J
v^{2} = 2(0.45 J)/(0.100 kg)
v = 3.0 m/s
Problem 2: Assume the height of the
roller coaster (see fig) is y = 40 m (take the reference point, y = 0,
as the bottom of the coaster hill). If the car starts from rest at point
A, what is the speed of the car at the bottom of the hill? What is the
speed of the car when it reaches point B?
Solution:
The energy at A is purely gravitational potential energy since the car
starts from rest. Therefore:
E_{A} = Mgy_{A} = M(9.80 m/s^{2})(40 m) = (392
M) J
The energy at the bottom of the hill is purely kinetic, no potential energy
since y = 0. Therefore
E_{f} = ½ Mv^{2 }=½Mv^{2}
Using conservation of energy, we can easily solve for v:
½Mv^{2} = (392 M) J
v^{2} =
2(392 )
v = 28.0 m/s
The energy at B is a combination of potential and kinetic energy, therefore:
E_{B} = ½ Mv^{2 }+ Mgy_{B} = ½
Mv^{2 }+ (245 M)
E_{B} = E_{A}
½ Mv^{2 }+ (245 M) = (392 M) where
v^{2} = 2(392 245)m^{2}/s^{2}
v^{2 }= 294 m^{2}/s^{2}
v = 17.1 m/s
Problem 3: A ball of mass 2.00kg
is dropped from a height of 1.5 m (from the ground) onto a massless spring
(the spring has an equilibrium length of 0.50 m). The ball compresses the
spring by an amount of 0.20 m by the time it comes to a stop. Calculate
the spring constant of the spring.
Solution:
First, we must pick a reference point. The solid line across the
diagram is our choice of y = 0. The initial energy of the system
is purely gravitational potential energy, and the final energy is a combination
of gravitational potential energy ( it is 0.20 m below our reference point)
and elastic potential energy (the object stops momentarily as it
fully compresses the spring). The mass is originally at a height
of 1.0 m above our reference point. Therefore:
mg(1.0 m) = mg(0.20 m) + ½ k(0.20 m)^{2}
(2.00 kg)(9.80 m/s^{2})(1.0 m) + (2.00 kg)(9.80 m/s^{2})(0.20
m) = 0.02 k
19.6 J + 3.92 J = 0.02 k
k = (23.5)/(0.02) = 1,180 N/m
PART 3: Conservation
of Energy and Nonconservative forces

A nonconservative force is a force that does
pathdependent work on an object, that is, to go from point A to B, it
matters what path is taken.

All dissipative or frictional forces are NONConservative.

Total Energy (includes mechanical energy and
other types of energy like thermal energy) is conserved under all circumstances,
even when nonconservative forces are involved.

The most general statement of conservation of
energy is: Energy can neither be destroyed nor created, energy is merely
transformed from one form to another.

For nonconservative forces: DE
= W_{nc}, the change in energy is equal to the work done by the
nonconservative force.
Problem 1: A sled of mass m is given
a kick on a frozen pond, imparting to it an initial speed v_{i }=
2.0 m/s. The coefficient of kinetic friction between the ice and the sled
is m_{k}=0.10.
Use energy considerations to find the distance the sled moves before stopping.
Solution:
The change in energy is equal to the work done by the nonconservative
force:
DE = E_{f}
 E_{i} = W_{nc}
W_{nc}= f_{k} d since the nonconservative force in this
case is friction, therefore
d = (E_{f}  E_{i})/(f_{k})
f_{k} = m_{k}mg
= (0.10)(mg)
The initial energy is purely kinetic and the final energy is 0:
E_{f}  E_{i} = 0  ½mv^{2}
d = (0  ½mv^{2})/((0.10)(mg))
= (2.0)/(0.98)
d = 2.04 m
Problem 2: Consider the same roller
coaster. It starts at a height of 40.0 m but once released, it can only
reach a height of 25.0 m above the reference point. If the mass of the
car is 1000kg, and the car traveled a distance of 400 m, estimate the
magnitude of the frictional force between the car and the track.
Solution:
W_{nc }= f_{k} d
f_{k}=(DE)/d
= (Ef + Ei)/ 400 m = (mgy_{b} + mgy_{a})/400 m
= (1000 kg)(9.80 m/s^{2})(40 m  25 m)/(400 m)
= 370 N
PART 4: Power

Power is the rate at which work is done

Power is a scalar quantity and has units called
WATTS or J/s

Average Power
Problem 1: A 700 N marine in basic
training climbs a 10.0 m vertical rope at a constant speed in a time of
8.0 s. What is his power output?
Solution:
In order for the marine to lift up the rope at a constant speed, he must
exert a force upward equal to his weight. The work he does is simple
his weight multiplied by his displacement:
W = (700 N)(10.0 m) = 7000 J
The power output is just this work divided by the time it took him to reach
the top:
P = W/t = (7000 J)/(8.0 s) = 875 W or about 880 W
Problem 2: A 200kg crate is pulled
along a level surface by an engine. The coefficient of friction between
the crate and the surface is m_{k}
= 0.40.
(a) How much power must the engine deliver
to move the crate at 5.0 m/s?
Solution:
The engine must work against the frictional force, therefore it has to
exert a force along the direction of displacement equal to the frictional
force. Therefore:
F = m_{k}mg
= (0.40)(200 kg)(9.80 m/s^{2}) = 784 N
P = F v = (784 N)(5.0 m/s) = 3,920 Watts
(b) How much work is done by the engine in
3 min. (3 min = 180 s)?
Solution:
P = W/t therefore, W = Pt
W = (3,920 Js)/(180 s) = 706 kJ
ADDITIONAL PROBLEM: A 220 kg crate
is lifted a distance of 21.0 m vertically with an acceleration of 1.47
m/s^{2} by a single cable.
(a) Find the tension on the cable
Solution:
The system is NOT in equilibrium since there is a net acceleration.
Therefore:
T  mg = ma
T = m (g + a) = (220 kg)( 9.80 + 1.47)
T = 2,480 N
(b) Find the net work done on the crate
Solution:
The net work done is done by the net force: F_{net} = ma = (220
kg)(1.47 m/s^{2}) = 323 N
W_{net} = (323 N)(21.0 m) = 6,800 J
(c) the work done by gravity on the crate
Solution:
W_{g} = (F_{g} cos 180)(21.0 m) = 45,300 J
(d) Find the work done by the cable on the
crate
Solution:
W_{T} = T d = (2480 N)(21.0 N) = 52,100 J
(e) find the final speed of the crate assuming
it started from rest.
Solution:
The werkenergy theorem states that the change in the kinetic energy of
the system is equal to the net wok done. The system starts from rest,
therefore the final kinetic energy = 0.
DK = 1/2(220
kg)v_{f}^{2}
1/2(220 kg)v_{f}^{2}= 6800 J
v_{f}^{2}= 2(6800 J)/(220 kg) = 61.8
v_{f}= 7,86 m/s
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