Homework 2 Flashcards
(63 cards)
1
Q
A newton is the force:
A) of gravity on a 1 kg body
B) of gravity on a 1 g body
C) that gives a 1 g body an acceleration of 1 cm/s2
D) that gives a 1 kg body an acceleration of 1 m/s2
E) that gives a 1 kg body an acceleration of 9.8 m/s
A
D) that gives a 1 kg body an acceleration of 1 m/s^2
2
Q
The inertia of a body tends to cause the body to: A) speed up B) slow down C) resist any change in its motion D) fall toward the Earth E) decelerate due to friction
A
C) resist any change in its motion
3
Q
Two objects, one having three times the mass of the other, are dropped from the same height in avacuum. At the end of their fall, their velocities are equal because:
A) anything falling in vacuum has constant velocity
B) both objects have the same acceleration
C) the acceleration of the larger object is three times greater than that of the smaller object
D) the force of gravity is the same for both objects
E) none of the above
A
B) both objects have the same acceleration
4
Q
The block shown moves with constant velocity on a horizontal surface. Two horizontal external forcesare applied to it, a 3 N force to the left and a 5 N force to the right, as shown. There is also a horizontalforce exerted by the surface of the block, which is not shown. The frictional force is: A) 0 B) 2 N, to the left C) 2 N, to the right D) slightly more than 2 N, to the right E) none of these
A
C) 2 N, to the left
5
Q
Equal forces~Fact on isolated bodies A and B. The mass of B is three times that of A. The magnitudeof the acceleration of A is: A) three times that of B B) 1/3 that of B C) the same as B D) nine times that of B E) 1/9 that of B
A
A) three times that of B
6
Q
A constant force of 10.0 N is exerted for 5.0 s on a 25-kg object, which was initially at rest. The changein speed of this object will be: A) 0.5 m/s B) 2 m/s C) 4 m/s D) 8 m/s E) 32 m/s
A
B) 2 m/s
7
Q
A ball with a weight of 1.0 N is thrown at an angle of 30◦above the horizontal with an initial speed of12 m/s. There is no air resistance acting on the ball. At its highest point, the net force on the ball is: A) 4.9 N, 30◦below horizontal B) zero C) 4.9 N, up D) 4.9 N, down E) none of these
A
E) none of these
8
Q
Two forces are applied to a 25 kg crate; one is 60.0 N to the north and the other is 30.0 N to the west.These are the only forces acting on the crate. The magnitude of the acceleration of the crate is: A) 0.50 m/s2 B) 1.5 m/s2 C) 2.7 m/s2 D) 5.2 m/s2 E) 11 m/s
A
C) 2.7 m/s^2
9
Q
A 1200 kg elevator is rising and its speed is increasing at 2.2 m/s2. The tension in the elevator cable is: A) 6000 N B) 8000 N C) 10000 N D) 12000 N E) none of these
A
E) none of these
14400 N
10
Q
A 1200 kg elevator is descending and its speed is increasing at 2.8 m/s2. The tension in the elevator cable is: A) 6400 N B) 8400 N C) 10000 N D) 14000 N E) None of these
A
B) 8400 N
11
Q
A pendulum bob weighing 2.5 N is held at an angleθfrom the vertical by a 5.0 N horizontal force~Fas shown. The tension in the string is supporting the pendulum bob (in newtons) is: A) 5 cosθ B) 2.5/sinθ C) 11 D) 5.6 E) none of these
A
E) none of these
12
Q
You stand on a spring scale on the floor of an elevator. Of the following, the scale shows the highestreading when the elevator: A) moves upward with increasing speed B) moves upward with decreasing speed C) remains stationary D) moves downward with increasing speed E) moves downward at constant speed
A
A) moves upward with increasing speed
13
Q
A crate of mass 50 kg is pushed across a frictionless horizontal floor with a force of 100 N, directed23.5◦below the horizontal. The magnitude of the normal force of the floor on the crate is: A) 130 N B) 180 N C) 340 N D) 530 N E) 610 N
A
D) 530 N
n=mg+Fsinθ= (50 kg)(9.8 m/s2) + (100 N)(sin 23.5◦) = 490 N + 39.9 N = 530 N
14
Q
A 27 N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline this 35◦ above the horizontal. The mass of the crate is: A) 4.3 kg B) 4.8 kg C) 5.7 kg D) 6.5 kg E) 160 kg
A
B) 4.8 kg
m=Fgsinθ=27 N(9.8 m/s2)(sin 35◦)=27 N(9.8 m/s2)(0.5736)= 4.8 kg.
15
Q
The “reaction” force does not cancel the “action” force because:
A) the action force is greater than the reaction force
B) the act on different bodies
C) they act in the same direction
D) the reaction force exists only after the action force is removed
E) the reaction force is greater than the action force
A
B) they act on different bodies
16
Q
A book rests on a table, exerting a downward force on the table. The reaction to this force is: A) the force of the Earth on the book B) the force of the table on the book C) the force of the Earth on the table D) the force of the book on the Earth E) the inertia of the book
A
B) the force of the table on the book
17
Q
A book rests on a table. Two force are exerted on the book: The downward force of the Earth (i.e.,gravity) on the book and the upward force of the table on the book. What is the reaction force to theforce of the Earth on the book?A) the force of the Earth on the book B) the force of the table on the book C) the force of the Earth on the table D) the force of the book on the Earth E) the inertia of the book
A
D) the force of the book on the Earth
18
Q
A 80 kg man stands in an elevator that has a downward acceleration of 1.3 m/s2. The magnitude ofthe force exerted by him on the floor is about: A) zero B) 90 N C) 560 N D) 680 N E) none of these
A
D) 680 N
19
Q
Two blocks are connected by a string and pulley as shown. Assuming that the string and pulley aremassless, what is the tension in the string? A) 0.27 N B) 0.77 N C) 0.97 N D) 1.1 N E) none of these
A
C) 0.97 N
T=2(0.090 kg)(0.110 kg)/(9.8 m/s2)(0.090 kg) + (0.110 kg)= 0.97 N.
20
Q
Referring to the same two blocks connected by a string and pulley in the previous problem, and stillignoring any possible friction, what is the magnitude of the acceleration of the blocks? A) 0.98 m/s2 B) 0.020 m/s2 C) 0.0098 m/s2 D) 0.54 m/s2 E) none of these
A
A) 0.98 m/s^2
a=(0.110 kg)−(0.090 kg)(0.090 kg) + (0.110 kg)(9.8 m/s2) = 0.98 N.
21
Q
Three books (X, Y, and Z) rest on a table. The weight of each book is indicated. The net force actingon book Z is: A) zero B) 9 N up C) 9 N down D) 19 N down E) none of these
A
A) zero
22
Q
Two blocks with massesmandMare pushed along a horizontal frictionless surface by a horizontalapplied force~Fas shown. The magnitude of the force of either of these blocks on the other is: A)mF/(m+M) B)mF/M C)mF/(M−m) D)MF/(M+m) E)MF/m
A
A) mF/(m+M)
23
Q
A brick slides on a horizontal surface. Which of the following will increase the frictional force on it?
A) Putting a second brick on top
B) Decreasing the surface area of contact
C) Increasing the surface area of contact
D) Decreasing the mass of the brick
E) None of the above
A
A) Putting a second brick on top
24
Q
Blocks weighing 70 N and 35 N are connected by a string as shown. Assuming the pulley is massless and the surface is frictionless, the tension in the string is: A) 18 N B) 23 N C) 30 N D) 35 N E) none of these
A
B) 23 N
T=m1m2/m1+m2g=(70 Ng)(35 Ng)/(70 Ng) + (35 Ng)g=(70 N)(35 N)/(70 N) + (35 N)(9.8 m/s2) = 23.3 N.
25
```
Referring to the same two blocks connected by a string and pulley in the previous problem, and stillignoring any possible friction, what is the magnitude of the acceleration of the blocks?
A) 3.3 m/s2
B) 3.9 m/s2
C) 4.9 m/s2
D) 6.7 m/s2
E) 9.8 m/s
```
A) 3.3 m/s^2
a=m2m1+m2g=(35 Ng)(70 Ng) + (35 Ng)g=35 N70 N + 35 Ng=13(9.8 m/s2) = 3.3 m/s2.
26
```
A forward horizontal force of 50 N is used to pull a crate, which weighs 250 N, at constant velocityacross a horizontal floor. The coefficient of kinetic friction is:
A) 0.5
B) 0.05
C) 2
D) 0.2
E) 20
```
D) 0.2
27
```
The speed of a hockey puck, which weighs 3.5 N, sliding across a level ice surface, decreases at a rateof 0.50 m/s2. The coefficient of kinetic friction between the puck and ice is:
A) 0.032
B) 0.041
C) 0.051
D) 0.12
E) none of these
```
μk=−ma/n=−(3.5 N/g)(−0.50 m/s2)/(3.5 N)=0.50 m/s2/9.8 m/s2= 0.051.
28
```
A block that weighs 50 N is initially at rest on a rough horizontal surface. A 20 N horizontal force isthen applied to the block. If the coefficients of friction areμs= 0.6 andμk= 0.3, the magnitude of the frictional force on the block is:
A) 8 N
B) 12 N
C) 18 N
D) 20 N
E) none of these
```
D) 20 N
29
```
For the block in the previous problem, the applied force is increased to 40 N. Now the magnitude ofthe frictional force on the block is:
A) 8 N
B) 12 N
C) 18 N
D) 20 N
E) none of these
```
E) none of these
30
```
A bread truck is traveling 18 m/s on a horizontal road. The brakes are applied and the truck skids toa stop in 3.6 s. The coefficienct of kinetic friction between the tires and road is:
A) 0.38
B) 0.51
C) 0.76
D) 0.92
E) none of these
```
B) 0.51
a=−(18 m/s)/(3.6 s) =−5.0 m/s2.
μk=−ma/mg=5.0 m/s2/9.8 m/s2= 0.51.
31
```
Three blocks (A, B, C), each having massM, are connected by strings as shown. Block C is pulled tothe right be a force,~Fthat causes the entire system to accelerate. Neglecting friction, the net forceacting on block B is:
A) zero
B)~F/3
C)~F/2
D) 2~F/3
E)~F
```
B) ~F/3
32
```
A block of massmis pulled along a rough horizontal floor by an applied force~Tas shown. The verticalcomponent of the force exerted on the block by the floor is:
A)mg
B)mg−Tcosθ
C)mg+Tcosθ
D)mg−Tsinθ
E)mg+Tsinθ
```
D)mg−Tsinθ
FNET, y= 0
=n+Tsinθ−mg
⇒n =mg−Tsinθ
33
```
A horizontal force of 15 N pushes a 0.50 kg book against a vertical wall. The book is initially at rest.Ifμs= 0.60 andμk= 0.40, the acceleration of the book is:
A) 0
B) 3.4 m/s2
C) 4.9 m/s2
D) 7.0 m/s2
E) none of these
```
A) 0
```
F−n= 0, soF=n.
fs,max=μsn= 0.6(15 N) = 9.0N.
mg= (0.5kg)(9.8 m/s2) = 4.9 N. So, the book does notmove, since there is an upward force of static friction of 4.9 N. The acceleration is zero.
```
34
```
In the same situation as the previous problem, the magnitude of the horizontal force is cut in half, to7.5 N. All else is the same. Now what is the magnitude of the acceleration of the book?
A) 0
B) 1.2 m/s2
C) 3.8 m/s2
D) 5.6 m/s2
E) none of these
```
C
Now the normal force is 7.5 N, so the maximum force of static friction is 4.5 N, which is less than theweight of the book. Hence, the book moves, and the frictional force has magnitudeμkn= 0.4(7.5 N) =3.0 N. Thus, the net downward force on the book is 1.9 N, and the acceleration isa=Fnet,y/m=(−1.9 N)/0.50 kg =−3.8 m/s2.
35
```
Block A, with a mass of 30 kg, rests on a 30◦incline. The coefficient of static friction is 0.20, and thecoefficient of kinetic friction is 0.10. The attached string is parallel to the incline and passes over amassless, frictionless pulley at the top. Block B, with a mass of 9.0 kg, is attached to the dangling endof the string. The acceleration of block A is:
A) 0.86 m/s2, up the plane
B) 0.86 m/s2, down the plane
C) 1.6 m/s2, up the plane
D) 1.6 m/s2, down the plane
E) 0
```
B) 0.86 m/s2, down the plane
36
```
An object of massmand another object of mass 2mare each forced to move along a circle of radius1.0 m at a constant speed of 1.0 m/s. The magnitudes of their accelerations are:
A) equal
B) in the ratio of√2 : 1
C) in the ratio of 2 : 1
D) in the ratio of 4 : 1
E) zero
```
A) equal
37
A car rounds a curve that follows the path of a circle with 100 m radius at a constant speed of 22 m/s.A ball is suspended by a string from the roof the car and moves with the car. The angle between thestring and the vertical direction is:
A) 19◦
B) 26◦
C) 34◦
D) 40◦
E) cannot be found without knowing the mass of the ball
B) 26
sinθcosθ= tanθ=v2/rg, so thatθ= arctan v2/rg = arctan(22 m/s)2/(100 m)(9.8 m/s2)= arctan 0.494 = 26.3◦.
38
```
One end of a 1.0 m long string is fixed, and the other end is attached to a 2.5 kg stone. The stone swings in a vertical circle near the Earth’s surface (sog= 9.8m/s2), passing the bottom point with speed 5.0 m/s. The tension force of the string at this point is about:
A) 29 N
B) 41 N
C) 52 N
D) 87 N
E) 98 N
```
D) 87 N
T=m(v2/r+g)= (2.5 kg)((5.0 m/s)2/1.0 m+ 9.8 m/s)= 87 N
39
```
A 10 kg crate is resting on a horizontal plank. The coefficient of static friction is 0.60 and the coefficientof kinetic friction is 0.40. After one end of the plank is raised so the plank makes an angle of 35◦withthe horizontal, the magnitude of the acceleration of the crate is:
A) 2.4 m/s2
B) 3.0 m/s2
C) 3.6 m/s2
D) 4.1 m/s2
E) none of these
```
A) 2.4 m/s^2
40
A force on a particle is conservative if:
A) its work equals the change in the kinetic energy of the particle
B) it obeys Newton’s second law
C) it obeys Newton’s third law
D) its work depends on the end points of the motion, not the path between
E) it is not a frictional force
D) Its work depends on the end points of the motion, not the path between
By definition a force on a particle is conservative if its work depends on the end points of the motion,not the path between.
41
```
A 20 kg object is moving at 5.0 m/s. A 3.0 N force is applied in the direction of motion and thenremoved after the object has traveled an additional 2.5 m. The work done by this force is:
A) 2.5 J
B) 5.0 J
C) 7.5 J
D) 10 J
E) 20 J
```
C) 7.5 J
42
An object moves in a circle at constant speed. The work done by the centripetal force is zero because:
A) the displacement for each revolution is zero
B) the average force for each revolution is zero
C) there is no friction
D) the magnitude of the acceleration is zero
E) the centripetal force is perpendicular to the velocity
E) the centripetal force is perpendicular to the velocity
43
The work done by gravity during the descent of a projectile:
A) is positive
B) is negative
C) is zero
D) depends for its sign on the direction of the y axis
E) depends for its sign on the direction of both the x and y axes
A) is positive
44
```
A sledge (including load) weighs 6000 N. It is pulled on level snow by a dog team exerting a horizontalforce on it. The coefficient of kinetic friction between sledge and snow is 0.03. How much work is doneby the dog team pulling the sledge for 500 m in a straight-line direction at constant speed?
A) 1.5×104J
B) 2.5×104J
C) 5.0×104J
D) 7.5×104J
E) 9.0×104J
```
E) 9.0x10^4
45
```
A 5.0 kg block is lifted vertically 0.7 m. The work done by gravity during this motion is about:
A) 34 J
B)−34 J
C) 17 J
D)−17 J
E) none of these
```
B) -34 J
46
```
A man pushes a crate which weighs 75 N upward along a frictionless slope that makes an angle of 30◦with the horizontal. His applied force is parallel to the slope. If the speed of the crate increases at arate of 0.5 m/s2, then the work done by the man while moving the crate a distance of 1.5 m is:
A)−62 J
B) 62 J
C)−41 J
D) 41 J
E) none of these
```
B) 62 J
Fman=ma+mgsinθ=(75 N/9.8 m/s2)(0.5 m/s2) + (75 N) sin 30◦= 41.3 N,andW= (41.3 N)(1.5 m) = 62 J.
47
```
Referring to the same situation as the previous problem, the work done by the force of gravity (duringthe same 1.5 m movement of the crate) is:
A)−56 J
B)−36 J
C)−16 J
D) 36 J
E) 56 J
```
A) -56 J
48
```
A block is attached to the end of an ideal spring and moved from coordinatexito coordinatexf. Therelaxed position is atx= 0. The work done by spring is negative if:
A)xi= 1 cm and xf= 3 cm
B)xi=−3 cm and xf= 2 cm
C)xi=−3 cm and xf= 0 cm
D)xi= 2 cm and xf=−1 cm
E)xi=−3 cm and xf= 2 cm
```
A)xi= 1 cm and xf= 3 cm
49
```
An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 80 N weight attached,the pointer indicates “40” on the scale as shown. Using a 120 N weight instead results in “50” on thescale. Using an unknown weight X instead results in “30” on the scale. The weight of X is:
A) 20 N
B) 30 N
C) 40 N
D) 50 N
E) 60 N
```
C) 40 N
50
```
When a certain rubber band is stretched a distancex, it exerts a restoring force of magnitudeF=Ax,whereA= 0.5 N/cm is a constant. The work done by a person in stretching this rubber band fromx= 0 tox= 1.0 cm is:'
A) 1.0 J
B) 0.25 J
C) 0.0025 J
D) 0.010 J
E) none of these
```
C) 0.0025 J
51
```
An ideal spring is hung vertically from the ceiling. When a 3.0 kg mass hangs at rest from it the springis extended 9.0 cm from its relaxed length. An upward external force is then applied to the block tomove it upward a distance of 12 cm. While the the block is being moved by the force, the work doneby the spring is
A)−1.2 J
B)−0.52 J
C)−0.26 J
D) 0.52 J
E) 1.2 J
```
E) 1.2 J
W=∫3.0 cm−9.0 cm−kydy=−12ky2∣∣∣∣3.0 cm−9.0 cm=12(3.27 N/cm)[81 cm2−9.0 cm2]= 118 N·cm = 1.2 J
52
Which of the following bodies has the largest kinetic energy?
A) Mass 9M with speedV
B) Mass 6M with speed 2V
C) Mass 3M with speed 3V
D) Mass M with speed 4V
E) All four of the above have the same kinetic energy
C) mass with 3M and speed 3V
53
```
A car with weight 7500 N is traveling at 20 m/s along a horizontal road when the brakes are applied.The car skids to a stop in 6.0 s. How much kinetic energy does the car lose in this time?
A) 3.8×104J
B) 8.5×104J
C) 1.5×105J
D) 4.2×105J
E) 4.8×106J
```
C) 1.5×105J
K=1/2mv2=1/2(7500 N/9.8 m/s2)(20 m/s)2= 1.5×105J
54
```
A crate of massmis initially at rest on a horizontal frictionless table. Then a constant horizontal forceFis applied. The crate begins moving in the +x-direction. What is the speed of the crate in terms ofthe force F and the displacement of the crate (i.e., the distance ∆xit has moved)?
A)√F∆x/m
B)√2F∆x/m
C) 2mF∆x
D)F∆x
E) none of these
```
B√2F∆x/m
W= ∆K, and since the crate is initially at rest (i.e., no initial kinetic energy),W=K=12mv2. SinceW=F∆x, the speed as a function of F and ∆x is: v=√2F∆x/m.
55
A 10 kg cart starts up an incline with a speed of 2.0 m/s and comes to rest 2.5 m up the incline. The total work done on the cart is:
A)−20 J
B)−12 J
C) 12 J
D) 20 J
E) impossible to calculate without more information
A) -20 J
The total (or net) work done on the cart is equal to the change of kinetic energy of the cart (W=∆K=Kf−Ki). Initial kinetic energy is ki=12mv2i=12(10 kg)(2.0 m/s)2= 20 J and final is Kf= 0 J, so the work done is−20 J.
56
```
A 2.5 kg block, which slides on a frictionless table, is attached to a horizonal ideal spring with a springconstant of 80 N/m. When the spring has its equilibrium length, the block has speed of 4.0 m/s. Whatis the maximum elongation of the spring?
A) 0.0
B) 0.50 m
C) 0.71 m
D) 1.0 m
E) none of these
```
C) 0.71 m
W=∫xm0(−kx)dx=−12kx2∣∣xm0=−12kx2m, xm=√−2Wk=√−2(−20 J)80 N/m= 0.71 m.
57
```
A 400 N force is the only force acting on a 5.0 kg object that starts from rest. At the instant the objecthas moved 2.5 m from its initial position, the rate at which the force is doing work (i.e., the power) is:
A) 50 W
B) 2.5 kW
C) 8.0 kW
D) 12 kW
E) 20 kW
```
C) 8.0 kW
P=~F·~v=Fv. The force is 400 N. We need the velocityvat the instant whenx= 2.5 m.v=at=Fmtandx=12at2=F2mt2, sov=at=Fm√2mxF=√2xFm=√2(2.5 m)(400 N)5.0 kg= 20 m/s.ThenP= (400 N)(20 m/s) = 8000 W = 8.0 kW.
58
```
An object is constrained by a cord to move in a circular path of radius 0.75 m on a horizontal frictionlesssurface. The cord will break if its tension exceeds 120 N. The maximum kinetic energy of the objectcan have is:
A) 8.0 J
B) 25 J
C) 45 J
D) 60 J
E) none of these
```
C) 45 J
The tension in the cord provides the centripetal force F=m v2/r. The maximum value it can provide is 120 N. Since K=1/2mv2,F=2K/r. Therefore, Kmax=r Fmax2=(0.75 m)(120 N)2= 45 J.
59
```
Referring to the same situation in the previous problem, what is the power provided by the tension in the cord?
A) zero
B) 8.0 W
C) 25 W
D) 45 W
E) none of these
```
A) zero
The force does no work, since its direction is perpendicular to the direction of motion. Since it doesno work, it provides no power:P= 0.
60
```
At timet= 0 a 2.0 kg particle has a velocity of (20 m/s)ˆi- (30 m/s)ˆj. Att= 4.0 s its velocity is(-20 m/s)ˆi+ (30 m/s)ˆj. During this time interval the total work done on the particle was:
A) zero
B)−1300 J
C) 2600 J
D) 1300 J
E) (40 J)ˆi+ (60 J)ˆ
```
A) zero
The speed of the mass is the same at both times, so its kinetic energy is the same. SinceW= ∆K,W= 0.
61
```
A person holds an object weighing 80 N at a height of 1.5 m above the floor for 60 seconds. The powerexpended while doing this is:
A) 120 W
B) 40 W
C) 2 W
D) 20 W
E) none of these
```
E) none of these
Since the object does not move, no work is done and no power is expended:P= 0.63.
62
```
A force does 18,000 J of work during a period of 2 hours. What is the average power?
A) 2.5 W
B) 4.0 W
C) 9.0 W
D) 25 W
E) none of these
```
A) 2.5 W
Pavg=W/∆t= (18,000J)/(7200 s) = 2.5W.
63
```
When a car accelerates, its engine must perform work in order to increase the car’s kinetic energy.Of course friction acting between the car’s moving parts and air resistance dissipate some energy, butignore these effects in this problem. If a particular car weighs 3620 lbs and can accelerate from 45 mphto 65 mph in 2.1 sec, then approximately what is the average power delivered by the engine during thisacceleration?
A) 110 hp
B) 150 hp
C) 190 hp
D) 230 hp
E) none of these
```
D) 230 hp
The work W done by the engine causes the change ∆K of kinetic energy of the car (W= ∆K via the work-energy theorem). The work is done during the time interval ∆t, so the average power isPave=W/∆t= ∆K/∆t.In this problem ∆t= 2.1 s, and we must calculate ∆K.The car’s massMis (3620 lbs)×(1 kg/2.2 lbs) = 1650 kg.Its initial speedvi= 45 mph×(0.447 ms−1/1 mph) = 20.1 m/s.Its final speedvf= 65 mph×(0.447 ms−1/1 mph) = 29.1 m/s.Then, ∆K=Kf−Ki=12Mv2f−12Mv2i= 0.5(1650 kg)[(29.1 m/s)2−(20.1 m/s)2] = 3.65×105J, soPave=∆K∆t=3.65×105J2.1 s= (1.74×105W)×1 hp746 W= 230 hp
