Transcript X – Work and Energy MC

Work and Energy
x
Work and Energy 06
Ref: Sec. 6.1
01
Work and Energy 06
A 500-kg elevator is pulled upward with a constant
force of 5500 N for a distance of 50.0 m. What is the
net work done on the elevator?
(A) 2.75 × 105 J
(B) -2.45 × 105 J
(C) 3.00 × 104 J
Work
W  F cos θ  x
(D) -5.20 × 105 J
Work and Energy 06
Ref: Sec. 6.1
02
Work and Energy 06
Matthew pulls his little sister Sarah in a sled on an icy
surface (assume no friction), with a force of 60.0 N at an
angle of 37.0° upward from the horizontal. If he pulls her
a distance of 12.0 m, what is the work done by Matthew?
(A) 185 J
(B) 433 J
Work
W  F cos θ  x
(C) 575 J
(D) 720 J
Work and Energy 06
Ref: Sec. 6.2
03
Work and Energy 06
A force moves an object in the
direction of the force. The graph
shows the force versus the object's
position. Find the work done when
the object moves from 0 to 6.0 m.
(A) 20 J
(B) 40 J
Work
(C) 60 J
Work is the product of
force times distance
(D) 80 J
Work and Energy 06
Ref: Sec. 6.3
04
Work and Energy 06
A horizontal force of 200 N is applied to move a 55-kg
cart (initially at rest) across a 10 m level surface. What
is the final kinetic energy of the cart?
(A) 1.0 × 103 J
(B) 2.0 × 103 J
(C) 2.7 × 103 J
(D) 4.0 × 103 J
Work/Energy Theorem
Wnet  ΔKE
Work
W  Fx
Work and Energy 06
Ref: Sec. 6.3
05
Work and Energy 06
If it takes 50 m to stop a car initially moving at 25 m/s,
what distance is required to stop a car moving at 50 m/s
under the same condition?
Work/Energy Theorem
(A) 50 m
Wnet  ΔKE
(B) 100 m
(C) 200 m
Work
(D) 400 m
W  Fx
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.3
06
Work and Energy 06
A spring-driven dart gun propels a 10-g dart. It is cocked by
exerting a constant force of 20 N over a distance of 5.0 cm.
With what speed will the dart leave the gun, assuming the
spring has negligible mass?
(A) 10 m/s
(B) 14 m/s
Work/Energy Theorem
Wnet  ΔKE
(C) 17 m/s
(D) 20 m/s
Work
W  Fx
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.3
07
Work and Energy 06
A 100-N force has a horizontal component of 80 N and a
vertical component of 60 N. The force is applied to a box
which rests on a level frictionless floor. The cart starts
from rest, and moves 2.0 m horizontally along the floor.
What is the cart's final kinetic energy?
(A) 200 J
(B) 160 J
Work
W  F cos θ  x
(C) 120 J
(D) zero
Work and Energy 06
Ref: Sec. 6.4-6.5
08
Work and Energy 06
A 15.0-kg object is moved from a height of 7.00 m above a
floor to a height of 15.0 m above the floor. What is the
change in gravitational potential energy?
(A) 1030 J
(B) 1176 J
(C) 1910 J
Gravitational Potential Energy
ΔPEg  mgΔy
(D) 2205
Work and Energy 06
Ref: Sec. 6.4-6.5
09
Work and Energy 06
A 400-N box is pushed up an inclined plane. The plane is 4.0 m
long and rises 2.0 m. If the plane is frictionless, how much work
was done by the push?
(A) 1600 J
(B) 800 J
(C) 400 J
Gravitational Potential Energy
W  ΔPEg  mgΔy
(D) 100 J
Work and Energy 06
Ref: Sec. 6.4-6.5
10
Work and Energy 06
A spring is characterized by a spring constant of 60 N/m.
How much potential energy does it store, when stretched
by 1.0 cm?
(A) 3.0 × 10-3 J
(B) 0.30 J
(C) 60 J
Spring Potential Energy
kx 2
PEs 
2
(D) 600 J
Work and Energy 06
Ref: Sec. 6.4-6.5
11
Work and Energy 06
A spring with a spring constant of 15 N/m is initially
compressed by 3.0 cm. How much work is required to
compress the spring an additional 4.0 cm?
(A) 0.0068 J
(B) 0.012 J
Spring Potential Energy
(C) 0.024 J
kx 2
PEs 
2
(D) 0.030 J
Work and Energy 06
Ref: Sec. 6.6-6.7
12
Work and Energy 06
A skier, of mass 40 kg, pushes off the top of a hill with an
initial speed of 4.0 m/s. Neglecting friction, how fast will
she be moving after dropping 10 m in elevation?
(A) 7.3 m/s
(B) 15 m/s
(C) 49 m/s
Conservation of Energy
ΔKE  ΔPEg  0
(D) 196 m/s
Work and Energy 06
Ref: Sec. 6.6-6.7
13
Work and Energy 06
A 1.0-kg ball falls to the floor. When it is 0.70 m above
the floor, its potential energy exactly equals its kinetic
energy. How fast is it moving?
(A) 3.7 m/s
(B) 6.9 m/s
Energy
PEg  KE
(C) 14 m/s
Potential Energy
(D) 45 m/s
PEg  mgy
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.6-6.7
14
Work and Energy 06
A roller coaster starts with a speed of 5.0 m/s at a point 45 m
above the bottom of a dip as shown in the diagram. Neglect
friction, what will be the speed of the roller coaster at the top
of the next slope, which is 30 m above the bottom of the dip?
(A) 12 m/s
Cons. of Energy
(B) 14 m/s
ΔKE  ΔPEg  0
(C) 16 m/s
Kinetic Energy
(D) 18 m/s
Potential Energy
PE  mgh
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.6-6.7
15
Work and Energy 06
A pendulum of length 50 cm is pulled 30 cm away from the
vertical axis and released from rest. What will be its speed
at the bottom of its swing?
(A) 0.50 m/s
(B) 0.79 m/s
(C) 1.2 m/s
(D) 1.4 m/s
Cons. of Energy
ΔKE  ΔPEg  0
Potential Energy
PE  mgh
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.6-6.7
16
Work and Energy 06
A 1500-kg car moving at 25 m/s hits an initially uncompressed
horizontal spring with spring constant of 2.0 × 106 N/m. What
is the maximum compression of the spring? (Neglect the mass
of the spring.)
Cons. of Energy
(A) 0.17 m
ΔKE  ΔPEg  0
(B) 0.34 m
(C) 0.51 m
(D) 0.68 m
Potential Energy
PE  mgh
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.8-6.9
17
Work and Energy 06
The kinetic friction force between a 60.0-kg object and a
horizontal surface is 50.0 N. If the initial speed of the object
is 25.0 m/s, what distance will it slide before coming to a stop?
(A) 15.0 m
Work/Energy Theorem
(B) 30.0 m
Wf  ΔKE
(C) 375 m
(D) 750 m
Work/Friction
Kinetic Energy
Wf  fk d
mv 2
KE 
2
Work and Energy 06
Ref: Sec. 6.8-6.9
18
Work and Energy 06
A force of 10 N is applied horizontally to a 2.0-kg mass on a
level surface. The coefficient of kinetic friction between the
mass and the surface is 0.20. If the mass is moved a distance
of 10 m, what is the change in its kinetic energy?
(A) 20 J
(B) 39 J
Work/Energy Theorem
ΔKE  Wnet
(C) 46 J
Work
(D) 61 J
W  Fx
Friction
fk  μ k N
Work and Energy 06
Ref: Sec. 6.10
19
Work and Energy 06
A 1500-kg car accelerates from 0 to 25 m/s in 7.0 s.
What is the average power delivered by the engine?
(1 hp = 746 W)
(A) 60 hp
Work/Energy Theorem
Wnet  ΔKE
(B) 70 hp
(C) 80 hp
Power
(D) 90 hp
W
P
t
Kinetic Energy
mv 2
KE 
2
Work and Energy 06
01
A 500-kg elevator is pulled upward with a constant
force of 5500 N for a distance of 50.0 m. What is the
net work done on the elevator?
F
Work
Wnet  Fnet x
Wnet  F  mg x


Wnet  5500 N  500 kg  9.8 m/s 2 50 m
mg
Wnet  30,000 J
Work and Energy 06
Matthew pulls his little sister Sarah in a sled on an icy
surface (assume no friction), with a force of 60.0 N at an
angle of 37.0° upward from the horizontal. If he pulls her
a distance of 12.0 m, what is the work done by Matthew?
Work
W  F cos θ  x
W  60 N cos 37  12 m
F
02
q
x
W  575 J
Work and Energy 06
03
A force moves an object in the
direction of the force. The graph
shows the force versus the object's
position. Find the work done when
the object moves from 0 to 6.0 m.
Work
Work is the product of
force times distance.
Work is equal to the area
under under the curve
20 J
40 J
20 J
80 J
Work and Energy 06
A horizontal force of 200 N is applied to move a 55-kg
cart (initially at rest) across a 10 m level surface. What
is the final kinetic energy of the cart?
Work
W  Fx
04
Work/Energy Theorem
Wnet  ΔKE
Fx  KE
KE  200 N10 m 
 2,000 J
Work and Energy 06
05
If it takes 50 m to stop a car initially moving at 25 m/s,
what distance is required to stop a car moving at 50 m/s
under the same condition?
Work/Energy Theorem
Work
W  Fx
Wnet  ΔKE
Kinetic Energy
 mv 2

Fx  
 0 
 2

2F v 2

 constant
m
x
mv 2
KE 
2
v12 v 22

x1 x 2
 v 22 
x 2  x1  
 v2 
 1
 50 m/s 
x 2  50m

 25 m/s 
2
x 2  200 m
Work and Energy 06
A spring-driven dart gun propels a 10-g dart. It is cocked by
exerting a constant force of 20 N over a distance of 5.0 cm.
With what speed will the dart leave the gun, assuming the
spring has negligible mass?
06
Work/Energy Theorem
Work
W  Fx
Wnet  ΔKE
 mv 2

Fx  
 0 
 2

Kinetic Energy
mv 2
KE 
2
2Fx  2 20 N  0.05 m
v
0.010 kg
m
 14.1 m/s
Work and Energy 06
07
A 100-N force has a horizontal component of 80 N and a
vertical component of 60 N. The force is applied to a box
which rests on a level frictionless floor. The cart starts
from rest, and moves 2.0 m horizontally along the floor.
What is the cart's final kinetic energy?
60
tan θ 
80
Work
 60 
W  F cos θ  x
θ  tan 1    36.9o
 80 
W  100 N cos 36.9  2.0 m
F
W  100 N cos 36.9  2.0 m
W  160 J
q
80
60
x
Work and Energy 06
A 15.0-kg object is moved from a height of 7.00 m above a
floor to a height of 15.0 m above the floor. What is the
change in gravitational potential energy?
Gravitational Potential Energy
ΔPEg  mgΔy

08
15 m

ΔPEg  15 kg 9.8 m/s 2 15 m  7 m 
7m
ΔPEg  1176 J
Work and Energy 06
09
A 400-N box is pushed up an inclined plane. The plane is 4.0 m
long and rises 2.0 m. If the plane is frictionless, how much work
was done by the push?
Gravitational Potential Energy
W  ΔPEg
W  mgΔy
W  400 N 2.0 m 
W  800 J
Work and Energy 06
A spring is characterized by a spring constant of 60 N/m.
How much potential energy does it store, when stretched
by 1.0 cm?
10
Spring Potential Energy
kx 2
PEs 
2
60 N/m 0.01 m 2
PEs 
2
PEs  0.003 J
Work and Energy 06
A spring with a spring constant of 15 N/m is initially
compressed by 3.0 cm. How much work is required to
compress the spring an additional 4.0 cm?
11
Spring Potential Energy
kx 2
PEs 
2
 kx 2 

ΔPEs  Δ
 2 

k 2
kx 22 kx12
 x 2  x12
ΔPEs 

2
2
2


15 N/m
0.07 m 2  0.03 m 2
ΔPEs 
2

 0.03 J
Work and Energy 06
A skier, of mass 40 kg, pushes off the top of a hill with an
initial speed of 4.0 m/s. Neglecting friction, how fast will
she be moving after dropping 10 m in elevation?
12
Conservation of Energy
ΔKE  ΔPEg  0
mv f2 mv i2

 mgy  0   0
2
2
v f  v i2  2gy
vf 
4.0 m/s 2  2 9.8 m/s 2 10 m
 14.6 m/s
Work and Energy 06
13
A 1.0-kg ball falls to the floor. When it is 0.70 m above
the floor, its potential energy exactly equals its kinetic
energy. How fast is it moving?
v
y
Kinetic Energy
Energy
Potential Energy
PEg  KE
PEg  mgy
mv 2
mgy 
2
mv 2
KE 
2
v  2gy


v  2 9.8 m/s 2 0.70 m
 3.7 m/s
Work and Energy 06
A roller coaster starts with a speed of 5.0 m/s at a
y
point 45 m above the bottom of a dip as shown in
the diagram. Neglect friction, what will be
y1
the speed of the roller coaster at the top of the
2
next slope, which is 30 m above the bottom of
the dip?
Cons. of Energy
Kinetic Energy
Potential Energy
ΔKE  ΔPEg  0
mv 2
KE 
PE  mgh
2
 mv 22 mv12 

  mgy 2  mgy1   0

 2

2


v 2  v12  2gy1  y 2  
14
5 m/s 2  2 9.8 m/s 2  45 m  30 m 
v 2  17.9 m/s
Work and Energy 06
15
A pendulum of length 50 cm is pulled 30 cm away from the
vertical axis and released from rest. What will be its speed
at the bottom of its swing?
Potential Energy
H
Cons. of Energy
PE  mgh
ΔKE  ΔPEg  0
 mv 2


 0   0  mgy   0
 2

x
Kinetic Energy
KE 
mv 2
y
2
v
H  L2  x 2
v  2gy

L

v  2 9.8m/s 2 0.10 m
v  1.4 m/s
H
0.5 m 2  0.3 m 2  0.40 m
y  LH
y  0.50 m  0.40 m  0.10 m
Work and Energy 06
A 1500-kg car moving at 25 m/s hits an initially uncompressed
horizontal spring with spring constant of 2.0 × 106 N/m. What
is the maximum compression of the spring? (Neglect the mass
of the spring.)
Potential Energy
PE  mgh
Cons. of Energy
16
Kinetic Energy
ΔKE  ΔPEg  0
mv 2
KE 
2


mv 2   kx 2
 0 
  
 0   0
2   2


mv 2
x
k
1500 kg 25 m/s 2

2.0 x 106 N/m
x  0.68 m
Work and Energy 06
The kinetic friction force between a 60.0-kg object and a
horizontal surface is 50.0 N. If the initial speed of the object
is 25.0 m/s, what distance will it slide before coming to a stop?
Work/Energy Theorem
Work/Friction
Wf  fk d
Wf  ΔKE
17
Kinetic Energy

mv 2 

 fk d   0 
2 

mv 2
KE 
2
2
mv 2


60
kg
25
m/s
d

2fk
2 50 N 
d  375 m
Work and Energy 06
18
A force of 10 N is applied horizontally to a 2.0-kg mass on a
level surface. The coefficient of kinetic friction between the
mass and the surface is 0.20. If the mass is moved a distance
of 10 m, what is the change in its kinetic energy?
Work/Energy Theorem
ΔKE  Wnet
Work
fk
W  Fx
ΔKE  WF  Wf
ΔKE  Fx   fk x 
ΔKE  Fx  μk mgx
Friction
F
x
fk  μ k N
fk  μk mg
ΔKE  10 N10 m   0.20 2 kg  9.8 m/s 2 10m 
ΔKE  60.8 J
Work and Energy 06
A 1500-kg car accelerates from 0 to 25 m/s in 7.0 s.
What is the average power delivered by the engine?
(1 hp = 746 W)
19
Work/Energy Theorem
Power
W
P
t
KE
P
t
Wnet  ΔKE
Kinetic Energy
mv 2
KE 
2
2
mv 2


1500
kg
25
m/s
P

2t
2 7.0 s 
P  89.8 hp
Work and Energy 06