Download AP 1st Qtr Exam Review Key

AP Physics
1st Qtr Exam Review
Name ________________________________________________ Date ____________
Topics to be considered:
I.
Motion along a Straight Line
A. Students should understand the general relationships among position, velocity, and acceleration for the motion of a
particle along a straight line so that
1. Given a graph of one of the kinematic quantities, position, velocity, or acceleration, as a function of time, they can
recognize in what time intervals the other two are positive, negative, or zero, and can identify or sketch a graph of
each as a function of time.
Describe the motion in the graphs below and make corresponding v vs. t & a vs. t graphs.
A: moving right, accelerating
B: moving left, accelerating
v
v
t
a
a
t
2.
t
Given an expression for one of the kinematic quantities, position, velocity, or acceleration, as a function of time,
they can determine the other two as a function of time, and find when these quantities are zero or achieve their
maximum and minimum values.
Find the velocity and acceleration functions for the motion described by
x = 5t 3 + sin 2t + 3
dx
= 15t 2 + 2 cos 2t
dt
dv
a=
= 30t − 4sin 2t
dt
v=
Find the position and velocity functions for the motion described by a
object is at x=0 and v=0 at t=0.
= 3cos t + 5(t 2 + 1) 2 . Assume that the
a = 3cos t + 5(t 4 + 2t 2 + 1) = 3cos t + 5t 4 + 10t 2 + 5
v = ∫ adt = ∫ 3cos t + 5t 4 + 10t 2 + 5dt
10 3
t + 5t + C
3
0 = 3sin 0 + 0 + 0 + 0 + C
C=0
10
v = 3sin t + t 5 + t 3 + 5t
3
10
x = ∫ vdt = ∫ 3sin t + t 5 + t 3 + 5tdt
3
6
4
t 10t
5t 2
x = −3cos t + +
+
+C
6 12
2
0 = −3cos 0 + 0 + 0 + 0 + C
C =3
v = 3sin t + t 5 +
x = −3cos t +
3.
t 6 10t 4 5t 2
+
+
+3
6 12
2
Students should know how to deal with situations in which acceleration is a specified function of velocity and time
so that they can write an appropriate differential equation and solve it for v(t), incorporating correctly a given initial
value of v.
Write a differential equation for an object whose acceleration is given by a
B.
=
− kv 2 dv − kv 2
.
=
m dt
m
Students should understand the special case of motion with constant acceleration so that they can:
1. Write down expressions for velocity and positions as function of time, and identify or sketch graphs of these
quantities.
2. Use the four kinematics equations to solve problems involving one dimensional motion with constant acceleration.
List the 4 kinematics equations:
v f = vi + at
v=
v f + vi
2
1
x f = xi + vit + at 2
2
2
2
v f = vi + 2a△ x
II.
Motion in Two & Three Dimensions
A. Students should know how to deal with displacement and velocity vectors so they can:
1. Relate velocity, displacement, and time for motion with constant velocity.
A motorist drives north for 35 min at 85 km/h and then stops for 15 min. She then continues north, traveling 130
km in 2.0 h. What is her total displacement?
 km 
 hr 
△ x =  85
 ( 35 min ) 
 + 130km = 180km, north
 hr 
 60 min 
What is her average velocity?
v=
2.
180km, north
= 63.4km / hr
 hr 
( 35 min + 15 min ) 
 + 2hour
 60 min 
Calculate the component of a vector along a specified axis, or resolve a vector into components along two
specified mutually perpendicular axes.
A displacement vector lying in the xy plane has a magnitude of 50.0 m and is directed at an angle of 120.0° to the
positive x axis. What are the rectangular components of this vector?
x = (50m) cos120° = −25.0m
y = (50m) sin120° = 43.3m
3.
Add vectors in order to find the net displacement of a particle that undergoes successive straight line
displacements.
While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75.0 m
north, 250 m east, 125 m at an angle 30.0° north of east, and 150 m south. Find the resultant displacement from
the cave entrance.
R
75
250
125
150
theta
90
0
30
270
x
0
250
108
0
358
y
75
0
62.5
-150
-12.5
R = 3582 + 12.52 = 358m
 −12.5 
 = −2.00°
 358 
358m, −2.00°
θ = tan −1 
4.
Subtract displacement vectors in order to find the location of one particle relative to another, or calculate the
average velocity of a particle.
A river has a steady speed of 0.500 m/s. A student swims upstream a distance of 1.00 km and returns to the
starting point. If the student can swim at a speed of 1.20 m/s in still water, how long does the trip take?
1000m
= 588s
1.2m / s + .5m / s
1000m
= 1429 s
tup =
1.2m / s − .5m / s
588s + 1429 s = 2017 s
tdown =
Compare this with the time the trip would take if the water were still.
2000m
= 1667 s
1.2m / s
1667 sec < 2017 sec
t=
5.
Add or subtract velocity vectors in order to calculate the velocity change or average acceleration of a particle, or
the velocity of one particle relative to another.
At t=0, a particle moving in the xy plane with constant acceleration has a velocity of (3.0i-2.0j) m/s at the origin. At
t=3.0 s, its velocity is given by v=(9.0i+7.0j)m/s. Find the acceleration of the particle.
a=
△v 9i + 7 j − (3i − 2 j ) 6i + 9 j
=
=
= (2.0i + 3.0 j )m / s 2
△t
3s
3s
Find the coordinates of the particle at any time t.
1
1
x = xi + vi t + at 2 = 0 + 3t + (2)t 2 = 3t + t 2
2
2
1
y = 0 − 2t + (3)t 2 = −2t + 1.5t 2
2
B.
Students should understand the general motion of a particle in two dimension so that, given functions x(t) and y(t)
which describe this motion, they can determine the components, magnitude and direction of the particle’s velocity and
acceleration as functions of times.
The position of a particle varies in time according to the expression r=(3.00i-6.00t2j) m. Find expressions for the
velocity and acceleration as functions of time.
v = (−12tj )m / s
a=(-12j)m/s 2
Determine the particle’s position and velocity at t=1.00 sec.
x(1sec) = (3.00i − 6.00 j )m
v(1sec) = −12.0 jm / s
C. Students should understand the motion of projectiles in a uniform gravitational field so they can
1. Write down expression for the horizontal and vertical components of velocity and position as functions of time, and
sketch or identify graphs of these components.
A soccer ball is kicked in the air. In the space below, make a sketch of the position vs. time graph, velocity vs.
time graph, and acceleration vs. time graph for the ball from the moment it leaves the ground until it returns.
2.
Use these expressions in analyzing the motion of a projectile that is projected above level ground with a specified
initial velocity.
A firefighter, 50.0 m away from a burning building, directs a stream of water from a fire hose at an angle of 30.0°
above the horizontal. If the initial speed of the water stream is 40.0 m/s, at what height does the water strike the
building?
vx ,i = 40 cos 30 = 34.6m / s
50m
= 1.44 s
34.6m / s
1
y f = yi + v y ,i t + at 2
2
v y ,i = 40 sin 30 = 20m / s
t=
1
2
y f = 0 + 20(1.44 s ) + (−9.8m / s 2 ) (1.44 s )
2
y f = 18.7 m
III. Force & Motion
A. Static Equilibrium
1. Students should be able to analyze situations in which a particle remains at rest, or moves with constant velocity,
under the influences of several forces.
The figure shows the same breadbox in four situations where horizontal forces are applied. Rank the situations
according to the magnitude of the box’s acceleration, greatest first.
a>b=c=d
B.
Dynamics of a Single Particle
1. Students should understand the relation between the force that acts on a body and the resulting change in the
body’s velocity so they can:
a. Calculate for a body moving in one direction, the velocity change that results when a constant force F acts
over a specified time interval.
Two horizontal forces act on a 5.0 kg chopping block that can slide over a frictionless kitchen counter, which
lies in an xy plane. One force is (2.0i+3.0j) N. Find the acceleration of the chopping block in unit vector
notation when the other force is (-3.0i-4.0j) N.
∑ F = ma
2i + 3 j + (−3i − 4 j ) = (5kg )a
a = (−0.20i − 0.2 j )m / s 2
b.
Calculate, for a body moving in one dimension, the velocity change that results when a force F(t) acts over a
specified time interval.
A force given by F=5.0t2+3.0t-1 acts on a body that is moving at 3.00 m/s at t=0. Find the velocity at t=4.00
sec. The object has a mass of 1.00 kg.
∑ F = 5t
+ 3t − 1
= 5t 2 + 3t − 1
1
m
5
3
v = ∫ adt = ∫ 5t 2 + 3t − 1dt = t 3 + t 2 − t + C
3
2
3= 0+0+0+C
a=
2
C =3
5
3
v = t3 + t2 − t + 3
3
2
5
3
2
v(4 sec) = (4)3 + (4) 2 − (4) + 3 = 129 m / s
3
2
3
c.
Determine, for a body moving in a plane whose velocity vector undergoes a specified change over a
specified time interval, the average force that acted on the body.
A 5.0 gram bullet leaves the muzzle of a rifle with a speed of 320 m/s. What average force is exerted on the
bullet while it is traveling down the 0.82 m long barrel of the rifle?
v 2f = vi2 + 2a△ x
(320m / s )2 = 0 + 2a (.82m)
a = 62, 439m / s 2
∑ F = ma = (.005kg )(62, 439m / s
2.
2
) = 312 N
Students should understand how Newton’s 2nd law applies to a body subject to forces, such as gravity, the pull of
strings, or contact forces, so they can
a. Draw a well-labeled free body diagram.
Draw and label a free body diagram for the mass, m, shown below.
N
T
mg
b.
Write down the vector equation that results from applying Newton’s 2nd law to the body, and take
components of this equation along appropriate axes.
For the diagram shown above, calculate the tension force and normal force if the mass is 30.0 kg and
θ=40.0°.
∑F
y
= ma y = 0
N = mg cos θ = (30kg )(9.8m / s 2 ) cos 40° = 225 N
∑F
x
= ma x = 0
T = mg sin θ = (30kg )(9.8m / s 2 )sin 40° = 189 N
3.
Students should be able to analyze situations in which a body moves with specified acceleration under the
influence of one or more forces so they can determine the magnitude and direction of the net force, or of one of
the forces that makes up the net force, in situations such as the following:
a. motion up or down with constant acceleration (in an elevator, for example).
Calculate the normal force on Einstein if the elevator accelerates upward at 5.00 m/s2 and his mass is 60.0
kg.
N − mg = ma
N
N = m( g + a ) = (60kg )(9.8m / s 2 + 5m / s 2 ) = 888 N
mg
b.
motion in a horizontal circle (e.g., mass on a rotating merry-go-round, or car rounding a banked curve).
The figure shows a conical pendulum that sweeps out a horizontal circle at a constant speed. If the bob has
a mass of 0.40 kg, the string is 0.90 m long, and the bob follows a circular path of circumference 0.9m What
are the tension in the string and the period of motion?
r = circumference / 2π = 0.9m /(2π ) = 0.143m
cos θ = r / L = .143m / .9m
θ = 80.8°
∑F
y
= ma y = 0
T sin θ = mg
mg
(.4kg )(9.8m / s 2 )
T=
=
= 3.97 N
sin θ
sin 80.8°
∑ Fx = ma x
2
 2π r 
m
2
period 
mv
4π 2 mr
T cos θ =
= 
=
r
r
period 2
period =
4π 2 mr
4π 2 (.4kg )(.143m)
= 1.89sec
=
T cos θ
3.97 N cos 80.8°
c.
Motion in a vertical circle (e.g., mass swinging on the end of a string, cart rolling down a curved track, rider
on a Ferris wheel).
A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At
the top of the hill, the normal force on the driver from the car seat is zero. The driver’s mass is 65.0 kg.
What is the magnitude of the normal force on the driver from the seat when the car passes through the
bottom of the valley?
at top:
∑ F = ma
mv 2
r
v = gr
mg =
at bottom:
∑ F = ma
mv 2
N − mg =
r
v2
gr
N = m( g + ) = m( g + ) = 2mg
r
r
4.
Students should understand the significance of the coefficient of friction so they can:
a. Write down the relationship between the normal and frictional forces on a surface.
Give the equation. f = µ N
b. Analyze situations in which a body slides down a rough inclined plane or is pulled or pushed across a rough
surface.
A loaded penguin sled weighing 75.0 N rests on a plane inclined at 40.0° to the horizontal. Between the sled
and the plane, µs=0.150 and µk=0.100. Find the smallest force that will prevent the sled from slipping, the
smallest force that will start the sled moving up the plane, and the force required to move the sled up the
plane at constant velocity.
smallest force to prevent slipping
∑F
y
= ma y = 0
N = mg cos θ
f = µ s N = µ s mg cos θ
∑F
x
= max = 0
F + f s − mg sin θ = 0
F = mg (sin θ − µ s cos θ ) = 39.6 N
smallest force to start moving up:
∑F
y
= ma y = 0
N = mg cos θ
f s = µ s N = µ s mg cos θ
∑F
x
= max = 0
F − f s − mg sin θ = 0
F = mg (sin θ + µ s cos θ ) = 56.8 N
smallest force to move at constant velocity:
∑F
y
= ma y = 0
N = mg cos θ
f k = µk N = µ k mg cos θ
∑F
x
= max = 0
F − f k − mg sin θ = 0
F = mg (sin θ + µk cos θ ) = 54.0 N
c.
5.
Analyze static situations involving friction to determine under what circumstances a body will start to slip, or
to calculate the magnitude of the force of static friction. above problem address this
Students should understand the effect of fluid friction on the motion of a body so they can:
a. Find the terminal velocity of a body moving vertically through a fluid that exerts a retarding force proportional
to velocity.
Assume that the resistive force exerted on a speed skater is f=-kmv2, where k is a constant and m is the
skater’s mass. The skater cross the finish line of a straight line race with a speed vf and then slows down by
coasting without skating. Show that the skater’s speed at any time t after crossing the finish line is
v(t ) =
vf
1 + ktv f
∑ F = ma
− kmv 2 = m
∫
t
t =0
dv
dt
−kdt = ∫
v (t )
vf
− kt = ∫
v (t )
vf
v −2 dv
− kt = −v −1
− kt = −
− kt −
dv
v2
v (t )
vf
1
1
+
v(t ) v f
1
1
=−
vf
v(t )
1
1
= kt +
v(t )
vf
v f kt + 1
1
=
v(t )
vf
v(t ) =
b.
vf
1 + ktv f
Describe qualitatively, with the aids of graphs, the acceleration, velocity, and displacement of such a particle
when it is released from rest or is projected vertically with specified initial velocity.
For the situation above, make a sketch of the x vs. t graph, v vs .t graph, and a vs. t graph.
vf
dv d 
a=
=  v(t ) =
dt dt 
1 + ktv f
C.
N
mg

−2kv f 2
−2
 = −2v f (1 + ktv f ) (kv f ) =
(1 + ktv f ) 2

Systems of Two or More Bodies
1. Students should understand Newton’s third Law so that, for a given force, they can identify the body on which the
reaction force acts and state the magnitude and direction of this reaction.
A slap bracelet rests on a table. Draw a freebody diagram of the slap bracelet. What is the reaction force to the
weight of the slap bracelet? The reaction force is the weight of the slap bracelet is the gravitational pull of the
slap bracelet on the Earth.
Is it possible for an action-reaction pair to be drawn on the same freebody diagram? yes or no (circle one)
The action and reaction pair act on different bodies, so they couldn’t be on the same diagram.
2.
Students should be able to apply Newton’s 3rd law in analyzing the force of contact between two bodies that
accelerate together along a horizontal or vertical line, or between two surfaces that slide across one another.
m2
T3
T1
If T4=90.0 N and m1=1.00 kg m2=4.00 kg, m3=2.00 kg, and m4=3.00 kg, find the acceleration of all the penguins
along with T1, T2, and T3,
a=
90 N
∑F =
= 9.00m / s
m
1
kg
2
kg
3
kg
4
kg
+
+
+
∑
summing forces on m1only:
T1 = m1a = (1kg )(9m / s ) = 9.00 N
2
N
T1
2
summing forces on m 2 only:
T2 − T1 = m2 a
m1 g
N
T1
T2
T2 = T1 + m2 a = 9 N + (4kg )(9m / s 2 ) = 45.0 N
m2 g
summing forces on m3only:
N
T2
T3 − T2 = m3 a
T3 = T2 + m3 a = 45 N + (2kg )(9m / s 2 ) = 63.0 N
3.
m3 g
Students should know that the tension is constant in a light string that passes over a massless pulley and should
be able to use this fact in analyzing the motion of a system of two bodies joined by a string.
A 2.5 kg and 7.5 kg mass hang from an Atwood’s machine. Find the acceleration of the masses and the tension
in the cord when the system is released from rest.
10 kg
2.5g
a=
2.5
kg
T3
7.5
kg
∑ F = (7.5kg )(9.8m / s
m
7.5g
) − (2.5kg )(9.8m / s 2 )
= 4.9m / s 2
7.5kg + 2.5kg
2
T
T − mg = ma
T = m( g + a ) = (2.5kg )(9.8m / s 2 + 4.9m / s 2 ) = 36.75 N
2.5
kg
mg
4.
Students should be able to solve problems in which applications of Newton’s laws lead to 2 or 3 simultaneous
linear equations involving unknown forces or accelerations.
If the mass shown is 20.0 kg, find the tension in all 3 cords.
T3 = (20kg )(9.8m / s 2 ) = 196 N
T2 cos 47° − T1 cos 28° = 0
T2 sin 47° + T1 sin 28° = 196
cos 47° − cos 28° T2
0
=
sin 47° sin 28° T1 196
T2 = 179 N and T1 = 138 N
f
IV. Energy
A. Work and the Work-Energy Theorem
1. Students should understand the definition of work so they can:
a. Calculate the work done by a specified constant force on a body that undergoes a specified displacement.
N
If you push a 40.0 kg crate at a constant speed of 1.40 m/s across a horizontal floor (µk=0.25), at what is
P
work being done on the crate by you and what energy is being dissipated by the friction force if it moves 2.8
m?
P = f = µ N = µ mg = 0.25(40kg )(9.8m / s 2 ) = 98 N
mg
WP = P△ x = (98 N )(2.8m) = 274.4 J
since v is constant, Wnet = 0 and Wf = −WP = −274.4 J
b.
Relate the work done by a force to the area under a graph of force as a function of position, and calculate
this work in the case where the force is a linear function of position.
Using the graph below, find the work done between x=0 m and x =12 m.
W = ∫ Fdx
1
1
(.008m)(12 N ) + (.002m)(6 N ) + (.002m)(6 N )
2
2
W = 0.066 J
W=
c.
Use integration to calculate the work performed by a force F(x) on a body that undergoes a specified
displacement in one dimension.
A can of peaches is made to move along an x axis from x=0.35 m to x=2.35 m by a force with a magnitude
given by F=5.0x3-2.0x-2. with x in meters and F in Newtons. How much work is done on the can by the force?
2.35 m
W = ∫ Fdx = ∫
2.35 m
.35 m
d.
5x 4 2
2
2
5 x − 2 x dx =
+
= 1.25(2.35) 4 +
− 1.25(.35) 4 −
= 33.2 J
4
x .35 m
2.35
.35
−2
3
Use the scalar product operation to calculate the work performed by a specified constant force F on a body
that undergoes a displacement in a plane.
A force F=(6.0i-2.0j) N acts on a particle that undergoes a displacement s = (3.0i+1.0j) m. Find the work
done by the force on the particle and the angle between F and s.
W = ∫ F • ds = (6i − 2 j ) • (3i + j ) = 6*3 − 2 = 16 J
a • b = a b cos Φ
16 =
(
6 2 + 22
)(
)
32 + 12 cos Φ
Φ = 36.9°
2.
Students should understand the work energy theorem so they can
a. State the theorem precisely, and prove it for the case of motion in one dimension.
Write it in equation form: W=∆K
b. Calculate the change in kinetic energy or speed that results from performing a specified amount of work on a
body.
A 0.700 kg particle has a speed of .200 m/s at point A and kinetic energy of 7.50 J at B.
1 2 1
mv = (.7 kg )(.2m / s ) 2 = 0.014 J
2
2
2K
2(7.5 J )
What is its speed at B? vB =
=
= 4.63m / s
m
.7 kg
What is its kinetic energy at A? K A
=
What is the total work done on the particle as it moves from A to B?
W = ∆K = K B − K A = 7.5 J − .014 J = 7.49 J
c.
Calculate the work performed by the net force, or by each of the forces that makes up the net force, on a
body that undergoes a specified change in speed or kinetic energy.
A 15.0 g bullet is accelerated in a rifle barrel 72.0 cm long to a speed of 780 m/s. Use the work-energy
theorem to find the average force exerted on the bullet while it is being accelerated.
W = ∆K
1
m(v 2f − vi2 )
2
mv 2f (.015kg )(780m / s ) 2
F=
=
= 6338 N
2∆x
2(.72m)
F ∆x =
d.
N
f
mg
Apply the theorem to determine the change in a body’s kinetic energy and speed that results from the
application of specified forces, or to determine the force that is required in order to bring a body to rest in a
specified distance.
A sled of mass m is given a kick on a frozen pond, imparting to it an initial speed vi=2.00 m/s. The coefficient
of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled
moves before stopping.
W = ∆K
1
m(v 2f − vi2 )
2
1
− µ mg ∆x = mvi2
2
2
v
∆x = i = 2.04m
2µ g
− f ∆x =
B.
Conservative Forces & Potential Energy
1. Students should understand the concept of a conservative force so they can:
a. State two alternative definitions of “conservative force” and explain why these definitions are equivalent.
Your definitions:
W ( x f → xi ) = −W ( xi → x f ) ?
Is the work a round trip?
Describe two examples each of conservative forces and non-conservative forces.
conservative: springs & gravity
non-conservative: friction and air resistance
Students should understand the concept of potential energy so they can:
a. State the general relation between force and potential energy, and explain why potential energy can be
associated only with conservative forces.
State the relation: F=-dU/dx
Why? because ∆U wouldn’t always vary indirectly with force
b. Calculate a potential energy function associated with a specified one dimensional force F(x).
A 3.00 kg block moving along the x axis is acted upon by a single force that varies with the block’s position
according to the equation F=ax2+b, where a=5.00 N/m2 and b=-2.50N. At x =1.0 m, the block is moving to
the right at 4.0 m/s. Determine its speed at x=2.0 m
b.
2.
F = 5 x 2 + 2.5
U = − ∫ Fdx = − ∫ 5 x 2 + 2.5dx = −
5 x3
+ 2.5 x + C
3
U i + Ki = U f + K f
5(1)3
1
5(2)3
1
+ 2.5(1) + C + (3kg )(4m / s ) 2 = −
+ 2.5(2) + C + (3kg )v f 2
3
2
3
2
v f = 4.7 m / s
−
c.
Given the potential energy function U(x) for a one-dimensional force, calculate the magnitude and direction of
the force.
The potential energy of a two particle system separated by a distance r is U(r) = A/r where A is a constant.
Find the force F in terms of A and r.
A
= Ar −1
r
dU
−2 A
F =−
= −2 Ar −2 = 2
dr
r
U=
d.
Write an expression for the force exerted by an ideal spring and for the potential energy stored in a stretched
or compressed spring.
F= =kx
U=
e.
1 2
kx
2
Calculate the potential energy of a single body in a uniform gravitational field.
A novice skier, starting from rest, slides down a frictionless 35.0° incline whose vertical height is 185 m. How
fast is she going when she reaches the bottom?
U i + Ki = U f + K f
mgh =
1 2
mv
2
v = 2 gh = 2(9.8m / s 2 )(185m) = 60.2m / s
f.
State the generalized work-energy theorem and use it to relate the work done by non-conservative forces on
a body to the changes in kinetic and potential energy of the body.
Two railroad cars, each of mass 7650 kg and traveling 95 km/h in opposite directions, collide head-on and
come to rest. How much thermal energy is produced in this collision?
2
1
1
 95, 000m 
6
∆Eint = ∆K = K f − K i = m(v 2f − vi2 ) = 0 − 2( )(7650kg ) 
 = 5.44 x10 J
2
2
3600
s


C. Conservation of Energy
1. Students should understand the concepts of mechanical energy and of total energy so they can:
a. State, prove, and apply the relation between the work performed on a body by non-conservative forces and
the change in a body’s mechanical energy.
An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a
snow bank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot’s mass
was 78 kg and his terminal velocity was 35 m/s, estimate the work done by the snow in bringing him to rest.
∆Eint = ∆K = K f − K i =
1
1
2
m(v 2f − vi2 ) = ( )(78kg ) ( 35m / s ) = 47, 775 J
2
2
Estimate the average force exerted on him by the snow to stop him.
F=
W 47, 775 J
=
= 43, 431J
∆x
1.1m
Estimate the work done on him by air resistance as he fell.
without drag:
v f2 = vi2 + 2a∆x
v f = 2a∆x = 2(9.8m / s 2 )(370m) = 85.2m / s
2
Wdrag = K no2 − K with
=
drag
b.
drag
1  2
 1 
2
2
m  vno − vwith
= m ( 85.2m / s ) − (35m / s )2  = 2.35 x105 J



2  drag drag  2
Describe and identify situations in which mechanical energy is converted to other forms of energy.
The masses of the javelin, discus, and shot are 0.80 kg, 2.0 kg, and 7.2 kg, respectively, and record throws
in the corresponding track events are about 98 m, 74 m, and 23 m, respectively. NAR, approximate the
minimum initial kinetic energies that would produce these throws.
Assume θ = 45°, so sin 2θ = 1
R=
v 2 sin 2θ v 2
=
g
g
v 2 = Rg
1 2 1
mv = mRg
2
2
1
K javelin = (.8kg )(98m)(9.8m / s 2 ) = 384 J
2
1
K disc = (2kg )(74m)(9.8m / s 2 ) = 725.2 J
2
1
K shot = (7.2kg )(23m)(9.8m / s 2 ) = 311.4 J
2
K=
Estimate the average force exerted on each object during the throw, assuming the force acts over a distance
of 2.0 m.
W ∆K
=
∆x ∆x
384 J
F javelin =
= 192 N
2m
725.2 J
Fdisc =
= 363 N
2m
811J
Fshot =
= 406 N
2m
F=
Do your results suggest that air resistance is an important factor? Explain.
Yes, air resistance must affect each of them very differently because the average force the person could
throw with should be the same.
c.
Analyze situations in which a body’s mechanical energy is changed by friction or by a specified externally
applied force.
In a needle biopsy, a narrow strip of tissue is extracted from a patient with a hollow needle. Rather than
being pushed by hand, to ensure a clean cut, the needle, can be fired into the patient’s body by a spring.
Assume the needle has a mass of 5.60 g, the light spring has force constant 375 N/m, and the spring is
originally compressed 8.10 cm to project the needle horizontally without friction. The tip of the needle then
moves through 2.40 cm of skin and soft tissue, which exerts a resistive force of 7.60 N on it. Next, the
needle cuts 3.50 cm into an organ, which exerts a backward force of 9.20 N on it. Find the maximum speed
of the needle.
U i + Ki = U f + K f
1 2 1 2
kxi = mv f
2
2
kxi2
(375 N / m)(0.081m) 2
vf =
=
= 20.96m / s
m
(0.0056kg )
Find the speed at which a flange on the back end of the needle runs into a stop, set to limit the penetration to
5.90 cm.
U i + Ki + W = U f + K f
1 2
1
kxi − f ∆x = mv 2f
2
2
1
1
(375 N / m)(0.081m)2 − (7.6 N )(0.024m) − (9.2 N )(−0.035m) = (0.0056kg )v 2f
2
2
v f = 16.1m / s
2.
Students should understand conservation of energy so they can:
a. Identify situations in which mechanical energy is or is not conserved.
b. Apply conservation of energy in analyzing the motion of bodies that are moving in a gravitational field and are
subject to constraints imposed by strings or surfaces. see above problems
c. Apply conservation of energy in analyzing the motion of bodies that move under the influence of springs.
A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k=4.00x102N/m. The
surface the block rests upon is frictionless. If the block is pulled out to xi=0.05 m and released, find the
speed of the block at xf=0.0250 m
U i + Ki = U f + K f
1 2 1 2 1 2
kxi = kx f + mv f
2
2
2
(400 N / m)(0.05m) 2 = (400 N / m)(0.025m)2 + (5kg )v 2f
v f = 0.387 m / s
Repeat the calculation but assume that the coefficient of kinetic friction is 0.150 between the surfaces.
U i + Ki + W f = U f + K f
1 2
1
1
kxi − µ mg ∆x = kx 2f + mv 2f
2
2
2
1
1
1
(400 N / m)(0.05m) 2 − .15(5kg )(9.8m / s 2 )(0.025m) = (400 N / m)(0.025m) 2 + (5kg )v 2f
2
2
2
v f = 0.277 m / s
d.
Apply conservation of energy in analyzing the motion of bodies that move under the influence of other
specified one dimensional forces.
A 2.1 x 103 car starts from rest at the top of a 5.0 m long driveway that is inclined at 20° with the horizontal.
If an average friction force of 4.0x103 impedes the motion, find the speed of the car at the bottom of the
driveway.
U i + Ki + W f = U f + K f
mgh − f ∆x =
1 2
mv f
2
(2100kg )(9.8m / s 2 )(1.71m) − (4000 N )(5m) =
1
(2100kg )v 2f
2
v f = 3.8m / s
3.
Students should be able to recognize and solve problems that call for application both of conservation of energy
and Newton’s Laws.
A Hot Wheels track is set up so that they car is started from the rest at the top of a hill. How high does the hill
have to be so that the car does not fall off the top of a circular loop with a radius of 15.0 cm?
∑ F =ma
mv 2
r
v = gr
mg =
h
mg
U i + Ki f = U f + K f
r
1
mgh = mg (2r ) + m( gr ) 2
2
1
5
h = 2r + r = r
2
2
D. Power
1. Students should understand the definition of power so they can:
a. Calculate the power required to maintain the motion of a body with constant acceleration (e.g., to move a
body along a level surface, to raise a body at a constant rate, or to overcome friction for a body that is
moving at a constant speed).
A 650 kg elevator starts from rest. It moves upward for 3.00 sec with constant acceleration until it reaches its
cruising speed, 1.75 m/s. What is the average power of the elevator motor during this period?
1
1
m ( v 2f − vi2 ) = (650kg )(1.75m / s ) 2 − 0 = 995 J
2
2
W 995 J
P=
=
= 332W
∆t
3s
W = ∆K =
How does this power compare with its power while moving at its cruising speed?
P = Fv = mgv = (600kg )(9.8m / s 2 )(1.75m / s ) = 10, 290 J
b.
Calculate the work performed by a force that supplies constant power, or the average power supplied by a
force that performs a specified amount of work.
A 200 kg crate is pulled along a level surface by an engine. The coefficient of friction between crate and
service is 0.40. How much power must the engine deliver to move the crate at 5.0 m/s?
push = µ mg = 0.4(200kg )(9.8m / s 2 ) = 784 N
P = F • v = (784 N )(5m / s ) = 3920W
How much work is done by the engine in 3.0 min?
W = ∫ Pdt = (3920 J / s )(3min)(60 s / min) = 705, 600 J
c.
Prove that the relation P=F•v follows from the definition of work, and apply this relation in analyzing particle
motion.
Start with
W = ∫ F • dx and P =
dW
and prove that P=F•v.
dt
dW = F • dx
P=
V.
dW F • dx
 dx 
=
= F •  = F •v
dt
dt
 dt 
Center of Mass
A. Students should understand the technique for finding center of mass so they can
1. Identify by inspection the center of mass of a body that has a point of symmetry.
Three uniform rods, each of length L=25 cm, form an inverted U. The vertical rods each have a mass of 18 kg.
The horizontal rod has a mass of 39 kg. Find the x and y coordinate of the center of mass.
(39kg )(12.5cm) + 18(0) + 18(25cm)
= 12.5cm
39kg + 18kg + 18kg
(39kg )(0) + 18(−12.5cm) + 18(−12.5cm)
=
= −6.00cm
39kg + 18kg + 18kg
xcm =
ycm
2.
Locate the center of mass of a system consisting of two such bodies.
A 3.00 kg particle is located on the x axis at x=-5.00 m, and a 4.00 kg particle is on the x axis at x=3.00 m. Find
the center of mass of this two particle system.
xcm =
(3kg )(−5.00m) + (4kg )(3.00m)
3
=− m
3kg + 4kg
7