14. semester exam review- fall 2015-solutions

8/18/2019 14. Semester Exam Review- Fall 2015-Solutions

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adame (eia275) – 14. Semester Exam Review: Fall 2015 – fackrell – (3.14)

This print-out should have 60 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 (part 1 of 2) 10.0 points

Consider a moving object whose position xis plotted as a function of the time t. Theobject moved in different ways during thetime intervals denoted I, II and III on thefigure.

2 4 6

2

4

6

t

x

I II III

During these three intervals, when was theobject’s speed highest? Do not confuse thespeed with the velocity.

1. During interval I

2. Same speed during intervals II and III

3. During interval II

4. Same speed during each of the three in-tervals.

5. During interval III correct

Explanation:The velocity v is the slope of the x(t) curve;

the magnitude v = |v| of this slope is thespeed. The curve is steepest (in absolute

magnitude) during the interval III and that iswhen the object had the highest speed.

002 (part 2 of 2) 10.0 pointsDuring which interval(s) did the object’s ve-locity remain constant?

1. During interval II only

2. During interval I only

3. During none of the three intervals

4. During each of the three intervals correct

5. During interval III onlyExplanation:

For each of the three intervals I, II or III, thx(t) curve is linear, so its slope (the velocityv) is constant. Between the intervals, thvelocity changed in an abrupt manner, but idid remain constant during each interval.

003 (part 1 of 2) 10.0 pointsConsider the following graph

0 2 4 6 8 10 12 14 16 18 2−10

−8−6

−4

−2

0

2

4

6

8

10

d i s p l a c e m e n t ( m )

time (s)

Displacement vs

Time

What is the displacement at 10 s?

1. −4 m correct

2. 2 m

3. −1 m

4. 1 m

5. −2 m

6. −3 m

7. Unable to determine


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8. 0 m

9. None of these

10. 3 m

Explanation:

Read the displacement from the graph.

004 (part 2 of 2) 10.0 pointsWhat is the velocity at 10 s?

1. −3 m/s

2. Unable to determine

3. None of these

4. 4 m/s

5. 3 m/s

6. 1 m/s

7. 0 m/s correct

8. −2 m/s

9. −1 m/s

10. 2 m/s

Explanation:At 10 s, the slope is 0.

005 (part 1 of 7) 10.0 pointsConsider the following graph of motion.

0 10 20 30 40 500

20

40

60

80

100

Time (sec)

D i s t a n

c e ( m )

S w i m

m e r

1

S w i m m e r

2

How many meters can Swimmer 1 cover in30 seconds?

1. 20 m

2. 80 m

3. 50 m

4. 10 m

5. 70 m

6. 100 m

7. 30 m

8. 40 m

9. 60 m correct

10. 90 m

Explanation:In the first 30 s, Swimmer 1 has moved from

0 m to 60 m.

006 (part 2 of 7) 10.0 pointsHow far can Swimmer 2 cover in 30 seconds?

1. 30 m correct

2. 80 m

3. 50 m

4. 100 m

5. 10 m

6. 70 m

7. 40 m

8. 20 m

9. 60 m

10. 90 m

Explanation:In the first 30 s, Swimmer 2 has moved from

0 m to 30 m.



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Predict the distance Swimmer 1 can go in 60seconds.

1. 80 m

2. 70 m

3. 50 m

4. 130 m

5. 40 m

6. 120 m correct

7. 100 m

8. 60 m

9. 110 m

10. 90 m

Explanation:There is an increase of 20 m every 10 s, so

the distance would be 120 m.

008 (part 4 of 7) 10.0 pointsPredict the distance Swimmer 2 can go in 60

seconds.

1. 50 m

2. 60 m correct

3. 120 m

4. 100 m

5. 80 m

6. 130 m

7. 70 m

8. 110 m

9. 40 m

10. 90 m

Explanation:There is an increase of 10 m every 10 s, s

the distance would be 60 m.

009 (part 5 of 7) 10.0 pointsWhich swimmer has the greatest speed?

1. The speeds are the same.

2. Swimmer 2

3. Unable to determine.

4. Swimmer 1 correct

Explanation:Swimmer 1 has the steeper graph, whic

means the greater speed.

010 (part 6 of 7) 10.0 pointsWhat is the speed of Swimmer 1?

1. 3 m/s

2. 5 m/s

3. 6 m/s

4. 2 m/s correct

5. None of these

6. 7 m/s

7. 8 m/s

8. 1 m/s

9. 4 m/s

Explanation:

speed = distance

time =

40 m

20 s = 2 m/s .

011 (part 7 of 7) 10.0 pointsWhat is the speed of Swimmer 2?

1. 8 m/s


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2. 4 m/s

3. 1 m/s correct

4. None of these

5. 5 m/s

6. 6 m/s

7. 3 m/s

8. 7 m/s

9. 2 m/s

Explanation:

speed = distance

time =

20 m

20 s = 1 m/s .

012 10.0 pointsThe distance traveled by an object per unittime is called

1. velocity.

2. acceleration.

3. speed. correct

4. momentum.

Explanation:

speed = distance

time .

013 10.0 points

Velocity is

1. the same as momentum.

2. the same as speed.

3. the same as acceleration.

4. speed in a specific direction. correct

Explanation:

014 (part 1 of 2) 10.0 pointsConsider a position-time graph.

The position is to be graphed

1. along either axis you choose.

2. along the vertical axis. correct

3. along the horizontal axis.

Explanation:The position is on the vertical axis and th

time on the horizontal axis.

015 (part 2 of 2) 10.0 pointsThe slope of the graph is

1. the speed.

2. the velocity. correct

3. the displacement.

4. the acceleration.

Explanation:

slope = position

time = velocity

016 10.0 pointsThe plot shows x(t) for a train moving alon

a long, straight track.

x

t

Which statement is correct about the motion?

1. The train moves at constant velocity.


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2. The train in fact moves at constant speedalong a circular path described by the x(t)curve.

3. The train continually speeds up.

4. The train at first speeds up, then slowsdown.

5. The train at first slows down, then speedsup.

6. The train continually slows down andcomes to rest. correct

Explanation:The slope of x(t) continually decreases to

reach zero. Since vx(t) = d x

dt ≈

∆ x

∆ t is by

definition the slope of x(t) at each t, the traincontinually slows down until it stops.

017 (part 1 of 3) 10.0 pointsConsider the plot below describing motionalong a straight line with an initial position of

x0 = 10 m.

−2

−1

0

1

2

3

1 2 3 4 5 6 7 8 9

time (s)

v e l o c i t y ( m / s )

What is the position at 2 seconds?

Correct answer: 13 m.

Explanation:The initial position given in the problem is

10 m.

1 2 3 4 5 6 7 8 0

1

2

3

−1−2

time (s)

v e l o c i t y ( m / s )

The position at 2 seconds is 10 meters pluthe area of the triangle (shaded in the abovplot)

x = 10 m + 1

2 (2 s − 0 s)

× (3 m/s − 0 m/s)

= 13 m ;

however, it can also be calculated:

x = xi + vi (tf − ti) + 1

2 (tf − ti)

2

= (10 m) + (0 m/s) (2 s − 0 s)

+ 1

2 (1.5 m/s2) (2 s − 0 s)2

= 13 m .

018 (part 2 of 3) 10.0 pointsWhat is the position at 6 seconds?

Correct answer: 21 m.

Explanation:The position is 13 m plus the area of th

trapezoid from 2 s to 6 s

x = 13 m + 1

2 (6 s − 2 s)

× (1 m/s + 3 m/s)

= 21 m ;

however, it can also be calculated:x = xi + vi (tf − ti) +

1

2 (tf − ti)

2

= (13 m) + (3 m/s) (6 s − 2 s)

+ 1

2 (−0.5 m/s2) (6 s − 2 s)2

= 21 m .

019 (part 3 of 3) 10.0 pointsWhat is the position at 8 seconds?


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Correct answer: 19.6667 m.

Explanation:The position is 21 m minus the area of the

triangle from 6 s to 8 s

x = (21 m) + 12

(8 s − 6 s)

× (−1.33333 m/s − 0 m/s)

= 19.6667 m ;

however, it can also be calculated

x = xi + vi (tf − ti) + 1

2 (tf − ti)

2

= 21 m + (0 m/s) (8 s − 6 s)

+ 1

2 (−0.666667 m/s2

) (8 s − 6 s)2

= 19.6667 m .

020 (part 1 of 2) 10.0 pointsThe initial speed of a body is 4.19 m/s.

What is its speed after 3.85 s if it acceleratesuniformly at 1.39 m/s2?

Correct answer: 9.5415 m/s.

Explanation:

Let : v0 = 4.19 m/s

a1 = 1.39 m/s2 , and

t = 3.85 s .

v = v0 + a1t ,

= 4.19 m/s + (1.39 m/s2) (3.85 s)

= 9.5415 m/s .

021 (part 2 of 2) 10.0 pointsWhat is its speed after 3.85 s if it acceleratesuniformly at −1.39 m/s2?


Explanation:

Let : a2 = −1.39 m/s2 .

v = v0 + a2 t

= 4.19 m/s + (−1.39 m/s

2

) (3.85 s)= −1.1615 m/s ,

which has a magnitude of 1.1615 m/s .

022 10.0 pointsA car, moving along a straight stretch of highway, begins to accelerate at 0.0142 m/s2. Itakes the car 27.2 s to cover 1 km.

How fast was the car going when it firsbegan to accelerate?


Explanation:

Let : a = 0.0142 m/s2 ,

t = 27.2 s , and

d = 1 km .

d = v0 t + 1

2 a t2

v0 = dt

− 12

a t

= 1 km

27.2 s −

1

2 (0.0142 m/s2)(27.2 s)

= 36.5716 m/s .

023 10.0 pointsIf the acceleration of an object is zero at sominstant in time, what can be said about itvelocity at that time?

1. It is positive.

2. It is not changing at that time. correct


4. It is negative.

5. It is zero.


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Explanation:The acceleration

a = ∆v

∆t = 0

∆v = 0 .

024 (part 1 of 2) 10.0 pointsA plane cruising at 269 m/s accelerates at13 m/s2 for 8.7 s.

What is its final velocity?


Explanation:

Let : v = 269 m/s ,

a = 13 m/s2 , andt = 8.7 s .

v = v0 + a t

= 269 m/s + (13 m/s2) (8.7 s)

= 382.1 m/s .

025 (part 2 of 2) 10.0 pointsHow far will it have traveled in that time?


Explanation:

sf = so + v0 t + 1

2 a t2

= v t + 1

2 a t2

= (269 m/s)(8.7 s) +

1

2 (13 m/s2

) (8.7 s)2

= 2832.28 m .

026 10.0 pointsIf you drop an object, it will accelerate down-ward at a rate of g = 9.8 m/s2.

If you throw it downward instead, its accel-eration (in the absence of air resistance) willbe

1. 9.8 m/s2 correct

2. greater than 9.8 m/s2.

3. less than 9.8 m/s2.


Explanation:The acceleration due to the gravity is in

dependent of any initial velocity and remainconstant.

027 10.0 pointsAn object is released from rest on a planethat has no atmosphere. The object fallfreely for 4.55 m in the first second.

What is the magnitude of the acceleratiodue to gravity on the planet?

Correct answer: 9.1 m/s2.

Explanation:

Let : s = 4.55 m .

s = 1

2 a t2

a = 2 s

t2 =

2 (4.55 m)

(1 s)2 = 9.1 m/s2 .

028 10.0 pointsA bullet is dropped into a river from a veryhigh bridge. At the same time another bulleis fired from a rifle aimed straight down at thriver.

Neglecting air resistance, how do the ac

celerations of the bullets compare just beforthey hit the water?

1. The bullet fired from the rifle does noaccelerate after leaving the barrel.

2. They are the same. correct

3. Acceleration is greatest for the bulledropped from rest.


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4. Acceleration is greatest for the bullet firedfrom the rifle.

5. We can’t answer unless we know how highthe bridge is.

6. We can’t answer unless we know how longit takes each bullet to hit the water.

Explanation:Both bullets are freely falling and so if air

resistance is negligible they both acceleratedownward at 10 m/s2.

029 (part 1 of 2) 10.0 pointsA small fish is dropped by a pelican that isrising steadily at 0.52 m/s.

After 3.5 s, what is the velocity of the fish?The acceleration of gravity is 9.81 m/s2.

Correct answer: −33.815 m/s.

Explanation:

Let : vi = 0.52 m/s ,

a = −9.81 m/s2 , and

∆t = 3.5 s .

The velocity of the fish is

vf = vi + a ∆t

= 0.52 m/s + (−9.81 m/s2)(3.5 s)

= −33.815 m/s .

030 (part 2 of 2) 10.0 pointsHow far below the pelican is the fish after the3.5 s?


Explanation:The new position of the fish is

∆xf = vavg ∆t = vi + vf

2 ∆t

= 0.52 m/s−33.815 m/s

2 (3.5 s)

= −58.2662 m

and the new position of the pelican is

∆x p = vavg∆t = (0.52 m/s)(3.5 s)

= 1.82 m ,

so the distance between the fish and the peli

can is

∆x p − ∆xf = 1.82 m − (−58.2662 m)

= 60.0862 m .

031 (part 1 of 3) 10.0 pointsA ball of mass 0.6 kg, initially at rest, ikicked directly toward a fence from a poin

20 m away, as shown below.The velocity of the ball as it leaves thkicker’s foot is 17 m/s at angle of 45 ◦ abovthe horizontal. The top of the fence is 6 mhigh. The ball hits nothing while in flight andair resistance is negligible.

The acceleration due to gravity is 9.8 m/s2

20 m

6 m 1 7

m / s

4 5 ◦

Determine the time it takes for the ball treach the plane of the fence.

Correct answer: 1.66378 s.

Explanation:

Let : θ = 45◦ and

d = 20 m .

The horizontal component of the velocity iconstant, so

vhoriz = v0 cos θ

= (17 m/s)cos45◦

= 12.0208 m/s .


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The horizontal motion defines the time of flight:

vhoriz t = d

t = d

vhoriz

= 20 m12.0208 m/s

= 1.66378 s .

032 (part 2 of 3) 10.0 pointsHow far above the top of fence will the ball

pass? Consider the diameter of the ball to benegligible.

Correct answer: 0.435986 m.Explanation:

The vertical component of the initial veloc-ity is

vvert = v0 sin θ

= (17 m/s) sin 45◦

= 12.0208 m/s .

The height of the ball during its flight is

given by

y = vvert t − 1

2 g t2

= (12.0208 m/s)(1.66378 s)

− 1

2 (9.8 m/s2) (1.66378 s)2

= 6.43599 m .

Therefore, the distance that the ball passesabove the fence is

∆y = (6.43599 m) − (6 m) = 0.435986 m .

033 (part 3 of 3) 10.0 pointsWhat is the vertical component of the velocitywhen the ball reaches the plane of the fence?

Correct answer: −4.28424 m/s.

Explanation:

Thus the the vertical component of the velocity when the ball reaches the plane of thfence is

vvert = v0 sin θ − g t

= (17 m/s) sin45◦

− (9.8 m/s2

) (1.66378 s)= −4.28424 m/s .

This is verified by analyzing the graph below

0 0.5 1.0 1.5 2.0 2−25

−20

−15

−10

−5

0

5

10

15

20

25

Time (s)

V e r t i c a l C o m p o n e n t o f V e l o c i t y ( m / s )

Vertical Velocity vs Time

034 10.0 pointsThe velocity of a projectile at launch has horizontal component vh and a vertical component vv. When the projectile is at the highest point of its trajectory, identify the verticaand the horizontal components of its velocityand the vertical component of its accelerationConsider air resistance to be negligible.

Vertical Horizontal VerticalVelocity Velocity Acceleration

1. 0 0 g

2. 0 vh 0

3. vv vh 0

4. vv 0 0

5. 0 vh g correct

Explanation:


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adame (eia275) – 14. Semester Exam Review: Fall 2015 – fackrell – (3.14) 1

The only force on the projectile is the grav-itational force, which gives the projectile aconstant vertical acceleration of magnitude g.

There is no acceleration in the horizontaldirection, which means at the highest point,the horizontal component of the velocity is

the same as the initial value vh.One other obvious thing: the vertical com-ponent of the velocity is zero at the highestpoint.

035 10.0 pointsA brick is thrown upward from the top of abuilding at an angle of 22.8◦ above the hori-zontal and with an initial speed of 14.3 m/s.

The acceleration of gravity is 9.8 m/s2 .If the brick is in flight for 3.2 s, how tall is

the building?


Explanation:Basic ConceptThe height of the building is determined by

the vertical motion with gravity acting downand an initial velocity acting upward:

y = y0 + v0y t − 1

2 g t2

SolutionChoose the origin at the base of the build-

ing. The initial position of the brick is y0 = h,the vertical component of the initial velocityis v0y = v0 sin θ directed upward, and y = 0when the brick reaches the ground, so

0 = h + v0y t − 1

2 g t2

h = −v0y t + 1

2 g t2

= −(5.54148 m/s)(3.2 s)

+ 1

2

9.8 m/s2

(3.2 s)2

= 32.4433 m .

036 10.0 pointsA ball is thrown horizontally from the top of a building 110 m high. The ball strikes the

ground 70 m horizontally from the point orelease.

What is the speed of the ball just before istrikes the ground?


Explanation:Basic Concepts:

vx = vx0 = constant

vy = vy0 − g t

x = vx0 t

y = vy0 t − 1

2g t2

Solution:

The vertical speed of the ball just before istrikes the ground is

vy =

2 g h

the flight time is

t =

2 h

g

so the horizontal speed of the ball is

vx = x

t = x

g

2 h

then the total speed just before it strikes thground is given by

v =

4.9x2

h + 19.6h

where we’ve used g = 9.8 m/s2.

037 (part 1 of 2) 10.0 pointsTom the cat is chasing Jerry the mouse acrosa table surface 0.7 m off the floor. Jerry stepout of the way at the last second, and Tomslides off the edge of the table at a speed o3.6 m/s.

Where will Tom strike the floor? The acceleration of gravity is 9.8 m/s2 .



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Explanation:

Let : vix = 3.6 m/s ,

viy = 0 m/s , and

y = −0.7 m .

His vertical motion defines the time. Sinceviy = 0 m/s,

y = viy t − 1

2 g t2 = −

1

2 g t2

t =

−2y

g =

−2(−0.7 m)

(9.8 m/s2)

= 0.377964 s .

The horizontal distance moved during thistime is

x = vix t = (3.6 m/s)(0.377964 s)

= 1.36067 m .

038 (part 2 of 2) 10.0 pointsWhat speed will Tom have just before he hits?


Explanation:The horizontal component of velocity does

not change during the flight, so Tom strikesthe floor with a horizontal component of ve-locity of

vx = vix = 3.6 m/s .

The vertical component of his velocity is

vy = viy − g t = −g t

= −

9.8 m/s2

(0.377964 s)

= −3.70405 m/s ,

so his total speed is

v =

v2x + v2y

=

(3.6 m/s)2 + (−3.70405 m/s)2

= 5.16527 m/s .


A book is at rest on an incline as shownbelow. A hand, in contact with the top othe book, produces a constant force F han

vertically downward.F hand

B o o k

The following figures show several attemptat drawing free-body diagrams for the book.

Which figure has the correct directions foeach force? The magnitudes of the forces arnot necessarily drawn to scale.

1.

weightforce

frictionnormal correct

2.

weightforce normal

friction

3.

normalfriction weight

force


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4.

weightfriction

forcenormal

5.

weightforce

normalfriction

6.

normal

force

frictionweight

7.

weightfriction

normalforce

8.

weightnormal

frictionforce

Explanation:The normal force points perpendicular to

the surface of the inclined plane. The weight

force points down. The F hand also pointsdown. The friction force keeps the book fromsliding and consequently points up the incline.

040 (part 2 of 2) 10.0 pointsFor the normal force exerted on the bookby the wedge in the diagram, which force(s)complete(s) the force pair for Newton’s thirdlaw (action-reaction)?

1. the component of F hand pointing perpendicular to the surface of the incline

2. the normal force exerted on the wedge bythe book correct

3. the component of gravity pointing paralleto the surface of the incline

4. the pull of the earth on the book

5. the sum of the component of gravity perpendicular to the surface of the incline andthe component of F hand perpendicular to thsurface of the incline

6. the pull of the book on the earth

7. the component of gravity pointing perpendicular to the surface of the incline

Explanation:The force that completes the third law pai

with the normal force of the wedge on thbook is the normal force of the book on thwedge.

The “the sum of the component of gravityperpendicular to the surface of the inclinand the component of F hand perpendiculato the surface of the incline” is equal anopposite to the normal force, but it is not action-reaction pair ; e.g., two forces equal andopposite to each other.

041 (part 1 of 2) 10.0 pointsFour forces act on an object.

F 1

F 4

F 2

F 3

If the object is accelerating to the rightcompare the forces.

1. F 1 = F 2, F 3 > F 4


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2. F 1 > F 2, F 3 < F 4

3. F 1 < F 2, F 3 < F 4

4. F 1 = F 2, F 3 < F 4 correct

5. F 1

= F 2

, F 3

= F 4

Explanation:The object is not accelerating up or down,

so F 1 = F 2. The object is accelerating to theright so F 4 is greater than F 3.

042 (part 2 of 2) 10.0 pointsIf the object is accelerating upward and to

the right, compare F 1 to F 2 and F 3 to F 4.

1. F 1 = F 2, F 3 = F 4

2. F 1 = F 2, F 3 < F 4

3. F 1 = F 2, F 3 > F 4

4. F 1 > F 2, F 3 < F 4 correct

5. F 1 < F 2, F 3 < F 4

Explanation:F 1 > F 2 causes acceleration upward and

F 4 > F 3 causes acceleration to the right.

043 10.0 pointsConsider the following system, where F =80 N, m = 2 kg, and M = 4 kg

M mF

What is the magnitude of the force withwhich one block acts on the other?

1. 80 N

2. None of these

3. 40 N

4. 26.6667 N correct

5. 160 N

6. 53.3333 N

Explanation:Let F contact be the magnitude of the force o

each box pushing on the other (by Newton’third law), and let right be positive.

The contact force supplies the accelerationon the rightmost box:

F contact = m a

a = F contact

m ,

so for the leftmost box

F − F contact = M a = M F contact

m

F contact = mM + m

F

= 2 kg

4 kg + 2 kg 80 N = 26.6667 N


A skier of mass 51.4 kg comes down a slopof constant angle 32◦ with the horizontal.

What is the force on the skier parallel to thslope? The acceleration of gravity is 9.8 m/s2

Correct answer: 266.931 N.

Explanation:

a

N W θ

The weight of the skier acting verticalldownward can be split into two componentsThe sine component acts along the slope.

W parallel = W sin θ = m g sin θ

= (51.4 kg) (9.8 m/s2)sin32◦

= 266.931 N.



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What force normal to the slope is exerted bythe skis?


Explanation:The cosine component acts perpendicular

to the slope.

W normal = W cos θ = m g cos θ

= (51.4 kg) (9.8 m/s2)cos32◦

= 427.179 N.

046 (part 1 of 2) 10.0 pointsAn elevator accelerates upward at 1.2 m/s2.

The acceleration of gravity is 9.8 m/s2 .What is the upward force exerted by the

floor of the elevator on a(n) 86 kg passenger?

Correct answer: 946 N.

Explanation:

N

a

mg

When the elevator is accelerating upward,

F net = m a = N − m g

N = m a + m g.

047 (part 2 of 2) 10.0 pointsIf the same elevator accelerates downwardswith an acceleration of 1.2 m/s2, what is theupward force exerted by the elevator floor onthe passenger?


Explanation:

N

a

mg

When the elevator is accelerating down-ward,

F net = m a = m g − N 2 N 2 = m g − m a.

048 10.0 pointsA 3.6 kg object hangs at one end of a rope thais attached to a support on a railroad boxcarWhen the car accelerates to the right, thrope makes an angle of 18◦ with the vertical

The acceleration of gravity is 9.8 m/s

2

.

a

3.6 kg

1 8 ◦

Find the acceleration of the car. (Hint

aobject = acar)


Explanation:

Given : m = 3.6 kg ,

θ = 18◦ , and

g = 9.8 m/s2 .

T

T sin θ

T cos θ θ

m g

VerticallyF y = T cos θ − m g = 0

T cos θ = m g . (1

Horizontally,

F x = T sin θ = m a . (2Dividing Eqs 1 and 2, we have

T sin θ

T cos θ =

a

g

tan θ = a

ga = g tan θ

=

9.8 m/s2

tan18◦

= 3.18421 m/s2 .


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049 (part 1 of 3) 10.0 pointsThree masses are connected by light stringsas shown in the figure.

m2

m3

m1

The string connecting the m1 and the m2passes over a light frictionless pulley.

Given m1 = 2.46 kg, m2 = 9.15 kg, m3 =4.29 kg, and g = 9.8 m/s2. The accelerationof gravity is 9.8 m/s2 .

Find the downward acceleration of m2mass.


Explanation:Consider the free body diagrams:

m2

m3

m1

T 1

a ↓

T 2

a ↓

T 1

↑ a

Applying Newton’s second law to each of these masses we get

m1 a = T 1 − m1 g (1)

m2 a = T 2 + m2 g − T 1 (2)

m3 a = m3 g − T 2 (3)

Adding these equations yields

(m1 + m2 + m3) a = (−m1 + m2 + m3) g ,

so

a =

−m1 + m2 + m3m1 + m2 + m3

g

=

−2.46 kg + 9.15 kg + 4.29 kg

2.46 kg + 9.15 kg + 4.29 kg

× (9.8 m/s2

)= 6.76755 m/s2 .

050 (part 2 of 3) 10.0 pointsFind the tension in the string connecting thm1 and the m2 masses.


Explanation:From equation (1),

T 1 = m1 (a + g)

= (2.46 kg)

6.76755 m/s2 + 9.8 m/s2

= 40.7562 N .

051 (part 3 of 3) 10.0 pointsFind the tension in the string connecting thm2 and the m3 masses.


Explanation:From equation (3),

T 2 = m3 (g − a)

= (4.29 kg)

9.8 m/s2 − 6.76755 m/s2

= 13.0092 N .

052 10.0 pointsAs viewed by a bystander, a rider in

“barrel of fun” at a carnival finds herself stuck

with her back to the wall.

ω


16/19


Which diagram correctly shows the forcesacting on her?

1. correct

2.

3.

4.

5. None of the other choices

6.

Explanation:The normal force exerted by the wall on

the rider provides the centripetal accelerationnecessary to keep her going around in a circle.The downward force of gravity is equal andopposite to the upward frictional force on her.

(Since this problem states that it is viewedby a bystander, we assume that the free-bodydiagrams are in an inertial frame.)

053 (part 1 of 2) 10.0 pointsCalculate the period of a ball tied to a string

of length 0.2 m making 3.5 revolutions everysecond.

Correct answer: 0.285714 s.

Explanation:

Let : f = 3.5 rev/s , and

R = 0.2 m .

T = 1

f =

1

3.5 rev/s = 0.285714 s .

054 (part 2 of 2) 10.0 pointsCalculate the speed of the ball.


Explanation:

v = D

T =

2 π R

T

= 2 π (0.2 m)

0.285714 s

= 4.39823 m/s .

055 10.0 points

Consider a conical pendulum, where string with length ℓ is attached to a masm. The angle between the string and the vertical is θ. The orbit is in the horizontal planwith radius r and tangential velocity v.

vr

g

ℓ

m

θ

What is the correct free body diagramshowing only the forces acting on the massThe acceleration due to gravity is 9.8 m/s2 .


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1.

θ

2.

θ

3.

θ

4.

θ

correct

5.

θ

Explanation:There are only two forces exerted on the

ball, the gravity and the tension on the string:

T

m g

θ

056 (part 1 of 2) 10.0 pointsAn air puck of mass 0.031 kg is tied to a stringand allowed to revolve in a circle of radius 1.4m on a frictionless horizontal surface. Theother end of the string passes through a holein the center of the surface, and a mass of 2.5 kg is tied to it, as shown. The suspended

mass remains in equilibrium while the puckrevolves on the surface.

1.4 m

2.5 kg

0.031 kg

What is the magnitude of the force thamaintains circular motion acting on the puckThe acceleration due to gravity is 9.81 m/s2


Explanation:

Let : g = 9.81 m/s2 ,

m p = 0.031 kg ,

m = 2.5 kg , and

r = 1.4 m .

r

m p

m

F c = F g = m g = (2.5 kg)(9.81 m/s2)

= 24.525 N .

057 (part 2 of 2) 10.0 pointsWhat is the linear speed of the puck?

Correct answer: 33.2803 m/s.Explanation:

F c = m pv2tr

vt =

F c r

m p=

(24.525 N) (1.4 m)

(0.031 kg)

= 33.2803 m/s .


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058 10.0 pointsIf a moon on Jupiter has 1/8 the mass of theEarth and 1/2 the Earth’s radius, what is theacceleration of gravity on the planet’s surface?The acceleration of gravity on Earth’s surface

is 10 m/s

2

.

1. 4 m/s2

2. 2 m/s2

3. 5 m/s2 correct

4. 3 m/s2

5. 1 m/s2

Explanation:

Let : M m = 1

8 M e ,

rm = 1

2 re , and

ge = 10 m/s2 .

The acceleration of gravity on a planet is

g =

G M

r2 .

Then,

gm =

G M mr2m

G M er2e

ge = 0.125

(0.5)2 (10 m/s2)

= 5 m/s2 .

059 10.0 pointsThe planet Mars has a mass of 6.1 × 1023 kgand radius of 3.2 × 106 m.

What is the acceleration of an objectin free fall near the surface of Mars?The value of the gravitational constant is6.67259 × 10−11 N · m2/kg2.


Explanation:

Let : M = 6.1 × 1023 kg ,

R = 3.2 × 106 m , and

G = 6.67259 × 10−11 N · m2/kg2 .

Near the surface of Mars, the gravitationforce on an object of mass m is

F = G M m

R2 ,

so the acceleration of an object in free fall is

a = F

m

= G M R2

= (6.67259 × 10−11 N · m2/kg2)

× 6.1 × 1023 kg

(3.2 × 106 m)2

= 3.97488 m/s2 .

060 10.0 pointsIf you moved to a planet that has twice th

mass of the Earth and also twice the diameterhow would your weight be affected?

1. It would be 8 times as much as it is now.



4. It would be the same as it is now.

5. It would be 14

as much as it is now.

6. It would be 1

6 as much as it is now.

7. It would be 1

2 as much as it is now

correct

Explanation:

On the Earth, g = G M E

r2 .


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On the planet, g ′ = G (2 M E )

(2 r)2 =

1

2 g , so

W P W E

= m g′

m g =

1

2 g

g =

1

2

W P = 12

W E .

The mass is doubled, so the direct rela-tionship for the mass gives twice the force(weight). The distance is also doubled, so theinverse square relationship for the distance

gives 1

4 the force (weight). The net result is

one half of the force (weight).

14. semester exam review- fall 2015-solutions

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