Work and kinetic energy | Bài tập môn Vật lý đại cương 1 CTTT | Trường Đại học Bách Khoa Hà Nội

02/07/2021 1 1 Kinetic energy 2 K  mv 2 K  kx (Spring) 2 2  2 2   2   
Work dW  Fds  W  dW   Fds  Fdscos(F,ds) 1 1 1
Chapter 6: WORK AND KINETIC ENERGY
Exercises: 1, 5, 7, 13, 19, 21, 25, 27, 29, 33, 35, 37, 39, 41, 47, 49, 53
Problems: 57(57), 63(62), 65(66), 69, 73, 77, 83, 85, 89, 91, 93, 97, 98, 99
Work-Kinetic Energy Theorem: Change in the kinetic energy of the particle = Net work done on the particle 1 1 2 2 W  K   K  K  mv  mv 2 1 2 2 2 2 Power  dW Fds  P    Fv Instantaneous power: dt dt Average power: W P  avg t  One Love. One Future. 1 One Love. One Future. 2 1 2
6.57(57). A luggage handler pulls a 20.0-kg suitcase up a ramp inclined at 25.0° above the horizontal by a force F of magnitude
140 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is k = 0.300. If the
suitcase travels 3.80 m along the ramp, calculate (a) the work done on the suitcase by the force F; (b) the work done on the
suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase
by the friction force; (e) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp,
what is its speed after it has traveled 3.80 m along the ramp?  
W  F.s  Fscos(F, s) F  (a) the work done by F  0
W  F.s  F.s.cos0 140 3.80  512J F (b) the work done by w   0
W  w.s  w.s.cos(115 )  20 9.8  0
1  3.80  cos115  315J w
(c) the work done by the normal force; W  n.s  0J n
(d) the work done by the friction force;    
W  f .s  n. scos(f ,s)   0 w cos 25   0 s cos(180 )  202J f k K
(e) the total work done on the suitcase W  W  W  W  W  15J total F f w n (f) The speed of the suitcase 1 2
W  K  K  mv  0  15  v  1.2 m / s total 2 1   2 One Love. One Future. 7/2/2021 Dang Duc Vuong - SEP - HUST 3 One Love. One Future. 4 3 4 1 02/07/2021
6.63(62). A 5.00-kg package slides 1.50 m down a long ramp that is inclined at 12.0° below the horizontal. The coefficient of kinetic friction
6.65(66). The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object
between the package and the ramp is k = 0.310. Calculate (a) the work done on the package by friction; (b) the work done on the
from the center of the earth. At the earth’s surface this force is equal to the object’s normal weight mg, where g = 9.8 m/s2,
package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the pack age. (e) If the package has
and at large distances, the force is zero. If a 20,000-kg asteroid falls to earth from a very great distance away, what will be its
a speed of 2.20 m/s at the top of the ramp, what is its speed after sliding 1.50 m down the ramp? (g) Find how far down the ramp the
package slides before it stop using the work-energy theorem.
minimum speed as it strikes the earth’s surface, and how much kinetic energy will it impart to our planet? You can ignore the     O  x : w sin   f  ma
effects of the earth’s atmosphere. x w  n  f  ma  n y O
 y : w cos  n  0 1 1 dr F   F  k. 2
(a) the work done by the friction force; f 1 mgR 2 r 2 2 r r E  F  k.  mg  k  mgR  F  M     2 E 2 R r 0 0 o
W  f .s  n. scos(f ,s)  w cos12    s  cos(180 )  2  2.3J r  R  F  mg E f  k   K  x E   12o 2 (b) the work done by w  0
W  w.s  w.s.cos(78 ) 15.3J  mgR dr 0 2 E     w dW F(r)dr dr cos180 mgR w F 2 E 2 r r
(c) the work done by the normal force; W  n.s  0J RE n RE RE  dr  1 2 2 W   dW   mgR   mgR    mgR
(d) the total work done on the suitcase W  W  W  W  7  .0J F F E 2 E E    r   r  total f w n  (e) The speed of the suitcase 1 1 2 2
W  K  K  mv  mv  7  .0  v 1.4 m / s 1 1 1 2 2 2
W  W  K  K  mv  mv  mv  mgR  v  2gR 11000 m / s total F 2 1 0 E E   total 2 1 0   2 2 2 2 2
(g) how far down the ramp the package slides before it stop 1 1 1
W  K K  mv  mv   5.0  0 2. 2 2 2 0 2  wscos78  0 wcos12   0
scos(180 ) s  2.6 m total 2 1 0 K   2 2 2 5 One Love. One Future. One Love. One Future. 6 5 6
6.69 A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless,
6.73 You and your bicycle have combined mass 80.0 kg. When you reach the base of a
horizontal surface (Fig.). The block is originally revolving at a distance of 0.40 m from the hole with a
bridge, you are traveling along the road at 5.00 m/s (Fig.). At the top of the bridge, you
speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the
have climbed a vertical distance of 5.20 m and have slowed to 1.50 m/s. You can
block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s. (a)
ignore work done by friction and any inefficiency in the bike or your legs. (a) What is
What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b)
the total work done on you and your bicycle when you go from the base to the top of
What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How
the bridge? (b) How much work have you done with the force you apply to the pedals?
much work was done by the person who pulled on the cord?
Let point 1 be at the base of the bridge and point 2 be at the top of the bridge
(a) The free-body diagram for the block 1 1 1 2 2
W  K  K  mv  mv  80.0  2 2 0.50 1.50  910 J total 2 1 2 1    2 2 2 2  v  W  W  W 2 2
F  ma  ma  m  mv 0.120 0.70 total person w x x rad  R   T    0.15N R 0.40 2 2    
W   dw   w ds 2 2
  w.ds.cos(w,ds)  wdscos  w.(0  h)  mgh  4  077 J w   F T    x 1 1 1 1 ds W  W  W  9  10  ( 4  077)  3170 J person total w   ds.cos 2 2 mv 0.120 2.80
(b) when the block has speed v = 2.80 m/s T    9.4N  R 0.10
(c) How much work was done by the person 2 2 1 1 0.120 2.80 0.120 0.70 w 2 2
W  W  K  K  mv  mv    0.44J total F 2 1 0 2 2 2 2 One Love. One Future. 7 One Love. One Future. 8 7 8 2 02/07/2021
6.77. A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing
6.83 Consider the system shown in Fig. The rope and pulley have negligible mass, and the
the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic
pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block
friction k = 0.30. Use the work-energy theorem to find how far the textbook moves from its initial position before coming to
is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving rest.
2.00 m. Use the work-energy theorem to calculate the coefficient of kinetic friction between
the 8.00-kg block and the tabletop.
Let point 1 be where the textbook is released and point 2 be where it stops sliding Apply W 1 1
tot= K2−K1 to the system consisting of both blocks A & B. 2 2
W  K  K  mv  mv  0  0  0 1 1 1 2 2 2 total 2 1 2 1 2 2 K  m v  m v  m  m v 1 A 1 B 1  A B  1 1 2 2 2 1 W  K  K   m  m v total 2 1  A B  2 2 W  W  W  kx  ( mgd) with x  0.250m  1 total spring friction k 2 K 0 2 2 1 2
 kx   mgd  0  d 1.1 m W  W  W  m gd   m gd k   2 total wA friction A k B 1 m m  m v
 m gd   m gd   m  m v      0.786 A k B  A B  A   2 2 A B 1 1 k 2 m 2m gd B B One Love. One Future. 9 One Love. One Future. 10 9 10
6.85. On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her
6.89 A physics student spends part of her day walking between classes or for recreation, during which time she expends
speed by (to) 45% due to a friction force that is 25% of her weight. Use the work-energy theorem to find the length of this
energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these rough patch.
activities, she expends energy at an average rate of 100 W. If she expends a total of 1.1x107 J of energy in a 24-hour day,
how much of the day did she spend walking?
Let point 1 be just before she reaches the rough patch and let point 2 be where she exits from the patch. 1 1
1 day = 24h = 24 x 3600 s = 8.64 x 104 s 2 2 W  K  K  mv  mv total 2 1 2 1 2 2 Let t 1
walk be the time she spends walking and tother be the time she spends in other activities v  3.0 m / s W  m 0.45v 2 2  v total 1 1  1   2
The energy expended in each activity is the power output times the time E  P  t v  45%v 2 1 7
1day : E  P  t  280 t 100 t  1.110 J ave walk other   W  W   mgd 4 t
 1.3110 s  218 min  3.6 h walk       total friction k 4 t  t  8.6410 s walk other   1 2 2
  mgd  mv (0.45 1)  d  1.5 m k 1   2 One Love. One Future. 11 One Love. One Future. 12 11 12 3 02/07/2021
6.91. The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is
6.93 Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and dis-
approximately 2000 MW. How many cubic meters of water must flow from the top of the dam per second to produce this
charges about 7500 l of blood. Assume that the work done by the heart is equal to the work required to lift this amount of
amount of power if 92% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of
blood a height equal to that of the average American woman (1.63 m). The density (mass per unit volume) of blood is water has a mass of 1000 kg.)
1.05x105 kg/m3. (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?
The power output is Pav = 2000 MW
(a) calculate m from the volume of blood pumped by the heart in one day
92% of the work done on the water by gravity is converted to electrical power output, so in m  V  
1.00 s the amount of work done on the water by gravity is 3 3  3 3       V  7500  7.50 m 1.05 10 kg.m 7.50 m 7.875 10 (kg) 3 m     P .t 200010 (1.00)  av  6  9 W    2.17410 J 0.92 0.92
One day, the work done by heart: 5
W  mgh  7.8759.811.63  1.2610 J
W = mgh so the mass of water flowing over the dam in 1.00 s must be (b) The heart's power output 9 W 2.17410 6 m   1.3010 kg 5 W 1.26 10 J gh 9.81170 P   1.46 w av 1  
t 24 h3600 s.h  6 The volume of water m 1.3010 kg 3 3 V   1.3010 m 3  1.0010  3 kg.m    One Love. One Future. 13 One Love. One Future. 14 13 14
6.97 Cycling. For a touring bicyclist the drag coefficient C(fair = 1/2.CAv2) is 1.00, the frontal area A is 0.463 m2 , and the coefficient
6.98. Automotive Power I. A truck engine transmits 28.0 kW (37.5 hp) to the driving wheels when the truck is traveling
of rolling friction is 0.0045. The rider has mass 50.0 kg, and her bike has mass 12.0 kg. (a) To maintain a speed of 12.0m/s (about 27
at a constant velocity of magnitude 60.0km/h (37.3 mi/h) on a level road. (a) What is the resisting force acting on the
mi/h) on a level road, what must the rider's power output to the rear wheel be? (b) For racing, the same rider uses a different bike with
truck? (b) Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. If the
coefficient of rolling friction 0.0030 and mass 9.00 kg. She also crouches down, reducing her drag coefficient to 0.88 and reducing her
force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed,
frontal area to 0.366 m2. What must her power output to the rear wheel be then to maintain a speed of 12.0 m/s? (c) For the situation in
what power will drive the truck at 30.0 km/h? At 120.0 km/h? Give your answers in kilowatts and in horsepower.
part (b), what power output is required to maintain a speed of 6.0 m/s? Note the great drop in power requirement when the speed is only
halved. (For more on aerodynamic speed limitations for a wide variety of human-powered vehicles, see "The Aerodynamics of Human-
(a) What is the resisting force?   
Powered Land Vehicles," Scientific American, December 1983.) Constant velocity F  F  F  0  F  F  F V= 60.0km/h net engine resis tance resis tance engine (a) on a level road P = ? A = 0.463 m2 3 P 28.010 (w) 1 3 3 P  Fv  F   1.6810 (N)  F  1.6810 (N) 1 1 1 resis tan ce1 P  F  v v  1000 m P = 28.0 kW 60.0 1 1 km.h     total 1 1 3600s 2
F  F  F   .N  CA v    m  m  2 g  CAv total roll air roll roll rider bike fair = 1/2.CAv2 2 2  P  513w
(b,c) The power of the truck when v2 = 30.0 km/h P  F  F v roll air  v  12.0m / s P v  v  30km / h 1 v  v  60km / h 2 1 F  0.65
  0.0045; m  50.0kg; m  12.0kg roll v  P 0.35P  roll rider bike 1 2 1 1 P  0.65  v v 10.3 kw  13.8 hp   2 3 2 2      1  P 0.35P 2 2   1 1 P 0.35P  v v
(b) For racing P = ? P  F  F  v  F ~ v  F  1 0.65  kv  k  1 1 2    air air    m  m  2 g  CA v  v  354 w 1 P 0.65 v v 1 1 3   roll air roll rider bike    2  Since F v v 3 air ~v2 and, 1 1  v v  1 1 v  v  120km / h  1 
P = Fv reducing the speed greatly 0.35P 0.35P 2 3 1 1
(c) v = 6.0 m/s P  F  F  v  k   F  v  m  m  2 g  CA v  v  52.1 w 3 air 3  P 0.35P  roll air roll rider bike    2 
 reduces the power required. v v 1 1 1 1 2 P  0.65  v v  114.8 kw 154 hp   2 3 3 3      v v  1 1 One Love. One Future. 15 One Love. One Future. 16 15 16 4 02/07/2021
6.99 Automotive Power II. (a) If 8.00 hp are required to drive a 1800-kg automobile at 60.0 km/h on a level road, what is the total retarding
A body of mass M (Fig. 1) with a small disc of mass m placed on it rests on a smooth
force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 km/h up a 10.00/0 grade (a hill rising 10.0
horizontal plane. The disc is set in motion in the horizontal direction with velocity v.
m vertically in 100.0 m horizontally)? (c) What power is necessary to drive the car at 60.0 km/h down a 1.00% grade? (d) Down what
To what height (relative to the initial level) will the disc rise after breaking off the
percent grade would the car coast at 60.0 km/h?  
body M? The friction is assumed to be absent. a) The total retarding force? P  (F  F )v 1 friction air 746 w P  8.00hp  8.00 hp  5968 w 1        1 hp P 5968 w 5968 w 1     F  F      358 N v  60.0kmh  1 resis tan ce 1000(m.km ) 16.67 1 1 m.s    tan  = 0.10  =5.71o 1 3600(s.h )
(b) v = 60.0 km/h, drive up, tan  = 0.10            F  F  F  F  W  N  0  F   F  F  W  N net engine friction air engine  friction air               P  F v  N  F
 F  W v  Nv  (F  F )v  Wv  (F  F )v  Wv
P  P  Wv  P  mgsin  .v 1 1   engine  friction air  friction air friction air   P  (F  F )v o
 5968 18009.81sin5.71  35168w  47.  1 hp 1 friction air
(c) v = 60.0 km/h, drive down, tan  = 0.010   = 0.5729o 
P  P  Wv  P  mgsin  o
.v  5968 1800 9.81 sin 0.5729  3028 w  4.06 hp 1 1    
(d) Down what percent grade would the car coast at 60.0 km/h? 
P  P  Wv  0  P  mgsin  .v  0  sin   0.02030  tan   0.0203  2.03% grade 1 1   One Love. One Future. 17 One Love. One Future. 18 17 18
A 1-kg block situated on a rough incline is connected to a spring of spring constant 100 N/m as shown in
figure. The block is released with initial velocity v = 3 m/s parallel to the incline with the spring in the K=100 N/m
unstretched position. The block moves 10 cm down the incline before coming to rest at point M. Assume
that the spring has a negligible mass and the pulley is frictionless. a) Find the coefficient of friction
between the block and the incline, b) How much time does it take for the block to travel? c) From M the M
block is slowly pulled down to point N, then released without initial speed. What is maximal value of
distance MN = s so the block stays at rest after releasing 370 Displacement along incline = x
Work Done by gravitational force =Wg =mgx sin37o
Work Done by frictional force =Wf =−μ mgx cos37o Workdone by al forces =ΔK.E=0 Wg +Wf +WS =0
mgx sin37o−μ mgx cos37o−21 kx2=0 Solving, μ=0.125 One Love. One Future. 19 One Love. One Future. 20 19 20 5 02/07/2021 www.hust.edu.vn
Thank you for your attentions!21 21 6