Download Displacement and Force in Two Dimensions Chapter 5

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CHAPTER 5
Supplemental Problems
DISPLACEMENT AND FORCE IN TWO DIMENSIONS
1. A small plane takes off and flies 12.0 km in a
direction southeast of the airport. At this
point, following the instructions of an air
traffic controller, the plane turns 20.0 to the
east of its original flight path and flies 21.0 km.
What is the magnitude of the plane’s
resultant displacement from the airport?
4. To get a cart to move, two farmers pull on
ropes attached to the cart, as shown below.
One farmer pulls with a force of 50.0 N in a
direction 35.0 east of north, while the other
exerts a force of 30.0 N in a direction 25.0
west of north. What are the magnitude and
the direction of the combined force exerted
on the cart?
2. A hammer slides down a roof that makes a
32.0 angle with the horizontal. What are
the magnitudes of the components of the
hammer’s velocity at the edge of the roof if it
is moving at a speed of 6.25 m/s?
5. Takashi trains for a race by rowing his canoe
on a lake. He starts by rowing along a straight
path. Then he turns and rows 260.0 m west. If
he then finds he is located 360.0 m exactly
north of his starting point, what was his
displacement along the straight path?
3. A worker has to move a 17.0-kg crate along a
flat floor in a warehouse. The coefficient of
kinetic friction between the crate and the
floor is 0.214. The worker pulls horizontally
on a rope attached to the crate, with a 49.0-N
force. What is the resultant acceleration of
the crate?
6. Mira received a 235-N sled for her birthday.
She takes the sled out to a flat field. When
she pushes it with a 45.0-N horizontal force, it
slides along at a constant speed. What is the
coefficient of kinetic friction between the
coefficient of kinetic friction between the sled
and the field?
Chapter 5  Displacement and Force in Two Dimensions
1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Name _____________________________
Supplemental Problems
7. A rod supports a 2.35-kg lamp, as shown
below.
continued
9. A child shoves a small toboggan weighing
100.0 N up a hill, giving the toboggan an
initial speed of 6.0 m/s. If the hill is inclined at
an angle of 32 above the horizontal, how far
along the hill will the toboggan slide? Assume
that the coefficient of sliding friction between
the toboggan and the hill is 0.15.
a. What is the magnitude of the tension in
the rod?
b. Calculate the components of the force
that the bracket exerts on the rod.
10. Two objects are connected by a string passing
over a frictionless, massless pulley. As shown
below, the block is on an inclined plane and
the ball is hanging over the top edge of the
plane. The block has a mass of 60.0 kg, and
the coefficient of kinetic friction between the
block and the inclined plane is 0.22. If the
block moves at a constant speed down the
incline, and the ball rises at a constant speed,
what is the mass of the hanging ball?
8. A 25.0-kg crate has an adjustable handle so
that it can be pushed or pulled by the handle
at various angles. Determine the acceleration
of the crate for each situation shown in the
diagram, given that the coefficient of sliding
friction between the floor and the bottom of
the crate is 0.20.
Chapter 5  Displacement and Force in Two Dimensions
2
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Supplemental Problems Teacher Support
DISPLACEMENT AND FORCE IN TWO DIMENSIONS
1.
A small plane takes off and flies 12.0 km in a direction southeast of the airport. At this point,
following the instructions of an air traffic controller, the plane turns 20.0 to the east of its original
flight path and flies 21.0 km. What is the magnitude of the plane’s resultant displacement from the
airport?
R2  A2  B2  2 AB cos 
R  (12.0 km)2  (21.0 km)2  2(12.0 km)(21.0 km)(cos 160.0)
 32.5 km
2.
A hammer slides down a roof that makes a 32.0 angle with the horizontal. What are the
magnitudes of the components of the hammer’s velocity at the edge of the roof if it is moving at a
speed of 6.25 m/s?
Fourth quadrant: vx . 0 and vy  0.
vx  v cos 
 (6.25 m/s)(cos  32.0)
 5.30 m/s
vy  v sin
 (6.25 m/s)(sin  32.0)
  3.31 m/s
Chapter 5  Displacement and Force in Two Dimensions
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Supplemental Problems Teacher Support
3.
continued
A worker has to move a 17.0-kg crate along a flat floor in a warehouse. The coefficient of kinetic
friction between the crate and the floor is 0.214. The worker pulls horizontally on a rope attached
to the crate, with a 49.0-N force. What is the resultant acceleration of the crate?
y-direction:
FN  Fg  mg
x-direction:
Fnet, x  Fp  Ff
 max  ma
Ff   k FN   k mg
ma  Fp   k mg
a
Fp   k mg
m
49.0 N  (0.214)(17.0 kg)(9.8 N/kg)

17.0 kg
2
 0.785 m/s
4.
To get a cart to move, two farmers pull on ropes attached to the cart, as shown below. One farmer
pulls with a force of 50.0 N in a direction 35.0° east of north, while the other exerts a force of 30.0
N in a direction 25.0° west of north. What are the magnitude and the direction of the combined
force exerted on the cart?
Rx  Ax  Bx  A cos 1  B cos 2
Ry  Ay  By  A sin 1  B sin 2
R  Rx2  Ry2
 ( A cos 1  B cos 2 )2  ( A sin 1  B sin 2 )2
((50.0 N) (cos 55.0)  (30.0 N)(cos 1150))2  ((50.0 N) (sin 550)  (30.0 N)(sin 115.0))2
 70.0 N
R 
 A sin 1  B sin2 
  tan1  y   tan1 

 Rx 
 A cos 1  B cos2 
 (50.0 N)(sin 55.0)  (30.0 N)(sin 115.0) 
 tan1 

 (50.0 N)(cos 55.0)  (30.0 N)(cos 115.0) 
 76.8
R  70.0 N at 76.8north of east
Chapter 5  Displacement and Force in Two Dimensions
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Supplemental Problems Teacher Support
5.
continued
a  0 (since v  constant)
Takashi trains for a race by rowing his canoe
on a lake. He starts by rowing along a straight
path. Then he turns and rows 260.0 m west. If
he then finds he is located 360.0 m exactly
north of his starting point, what was his
displacement along the straight path?
Fp  Ff  
Fp  Ff
 k FN  k Fg
k 
Fp
Fg
45.0 N
235 N
 0.191

7.
A rod supports a 2.35-kg lamp, as shown
below.
Rx  0.0 m, Ry  360.0 m
Bx  260.0 m, By  0.0 m
R  A  B, A  R  B
Ax  Rx  Bx  0.0 m  (260.0 m)
a. What is the magnitude of the tension in
the rod?
y - direction :
Ty  Fg
 260.0 m
Ay  Ry  By  360.0 m  0.0 m
 360.0 m
A  Ax2  Ay2

T sin   mg
(260.0 m)2  (360.0 m)2
mg
sin 
(2.35 Kg)(9.8 N/kg)

sin 25.0
 54 N
T
 444.1 m
 Ay 

 Ax 
 360.0 m 
 tan1 
  54.16
 260.0 m 
A  444.1 m at 54.16 north of east
  tan1 
6.
b. Calculate the components of the force
that the bracket exerts on the rod.
x - direction:
Tx  Fx  0
Fx  Tcos 
 (54 N)(cos 25.0)
Fx  49 N, inward
y - direction:
Mira received a 235-N sled for her birthday.
She takes the sled out to a flat field. When
she pushes it with a 45.0-N horizontal force,
it slides along at a constant speed. What is
the coefficient of kinetic friction between
the sled and the field?
Fy  Ty
y -direction: FN  Fg
 mg
 (2.35 kg) (9.8 N/kg)
Fy  23.0 N, upward
x -direction: Fp  Ff  ma
Chapter 5  Displacement and Force in Two Dimensions
5
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Supplemental Problems Teacher Support
8.
continued
A 25.0-kg crate has an adjustable handle so
that it can be pushed or pulled by the handle
at various angles. Determine the
acceleration of the crate for each situation
shown in the diagram, given that the
coefficient of sliding friction between the
floor and the bottom of the crate is 0.20.
For (a)
y -direction :
FN  Fg  mg
FN  mg
x -direction :
Fnet, x  Fp  Ff
Fnet, x  max  ma
ma  Fp  Ff
Fp  F cos 
Ff  k FN  k mg
Ff  k FN  k (mg  F sin  )
ma  Fp  k mg
a
Fp  k mg

Fp
Fnet, x  F cos   k (mg  F sin  )  ma
 k g
Fnet
m
F cos   k (mg  F sin  )

m
F cos   k F sin  )

 k g
m
(70.0 N)(cos 30.0 )  (0.20)(70.0 N)(sin 30.0 )

25.0 kg
 (0.20)(9.8 N/kg)
a
m
m
70.0 N

 (0.20)(9.8 N/kg)
25.0 kg
 0.84 m/s2
For (b)
y -direction :
FN  Fg  Fy
FN  mg  F sin 
 0.18 m/s2
x -direction :
Fnet, x  Fp  Ff
Fnet, x  max  ma
ma  Fp  Ff
Chapter 5  Displacement and Force in Two Dimensions
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Supplemental Problems Teacher Support
For (c)
y -direction :
FN  Fg  Fy
FN  mg  F sin 
x -direction :
Fnet, x  Fp  Ff
Fnet, x  max  ma
ma  Fp  Ff
Fp  F cos 
Ff   k FN   k (mg  F sin  )
Fnet, x  F cos   k (mg  F sin  )  ma
Fnet
m
F cos   k (mg  F sin  )

m
F cos   k F sin  )

 k g
m
(70.0 N)(cos 30.0)  (0.20)(70.0 N)(sin 30.0)

 (0.20)(9.8 N/kg)
25.0 kg
 0.74 m/s2
a
Chapter 5  Displacement and Force in Two Dimensions
7
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continued
Supplemental Problems Teacher Support
9.
continued
A child shoves a small toboggan weighing 100.0 N up a hill, giving the toboggan an initial speed of
6.0 m/s. If the hill is inclined at an angle of 32° above the horizontal, how far along the hill will the
toboggan slide? Assume that the coefficient of sliding friction between the toboggan and the hill is
0.15.
y -direction :
Fnet, y  may  0
FN  Fgy  0
FN  Fgy  mg cos 
x -direction :
Fnet, x  max  ma
Fgx  mg sin 
Fgx  Ff  ma
F  Ff  ma
Ff  k FN  k mg cos 
mg sin   k mg cos   ma
a  g(sin   k cos  )
vf2  vi2  2a(xf  xi )
0  vi2  2ax
d
vi2
vi2

2a 2g(sin  k cos  )
(6.0 m/s)2
(2)(9.8 N/kg)(sin 32  (0.15) cos 32.0)
 4.6 m, up the hill

10. Two objects are connected by a string passing over a frictionless, massless pulley. As shown below,
the block is on an inclined plane and the ball is hanging over the top edge of the plane. The block
has a mass of 60.0 kg, and the coefficient of kinetic friction between the block and the inclined
plane is 0.22. If the block moves at a constant speed down the incline, and the ball rises at a
constant speed, what is the mass of the hanging ball?
Chapter 5  Displacement and Force in Two Dimensions
8
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Supplemental Problems Teacher Support
v is constant, so there is no acceleration.
Ball (m2 ) :
y - direction :
Fnet, y  m2 ay  0
FT  Fg  0
FT  m2 g
Block (m1 ):
y - direction : (perpendicular to the incline)
Fnet, y  m1 ay  0
FN  Fgy  0
FN  Fgy  m1 g cos 
Ff  k FN  k m1 g cos 
x - direction : (parallel to the incline)
Fnet, X  m1 aX  0
Fgx  Ff  FT  0
FgX  m1 g sin 
m1g sin   k m1g cos   FT  0
m1g sin   k m1g cos   m2 g  0
m g sin   k m1g cos 
m2  1
g
 m1 sin   k m1 cos 
 (60.0 kg)(sin 35.0)  (0.22)(60.0 kg)(cos 35.0)
 24 kg
Chapter 5  Displacement and Force in Two Dimensions
9
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
continued