Tài liệu Physics exercises_solution: Chapter 06 docx

c The normal force is mgFsin, and so the work done by friction is J5.386 30sinN2.99s/m80.9 kg30 force and gravity act perpendicular to the direction of motion, so neither force does wo

6.1: a)(2.40N (1.5m)3.60J b) (0.600N)(1.50m)0.900J

c) 3.60J0.720J2.70J

6.2: a) “Pulling slowly” can be taken to mean that the bucket rises at constant speed, so

the tension in the rope may be taken to be the bucket’s weight In pulling a given length

of rope, from Eq (6.1),

J

6.264)m00.4)(

s/m80.9(kg75.6





Fs mgs W

b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq (6.2) gives the negative of the result of part (a), or 265J c) The net work done on the bucket

is zero

6.3: (25.0N)(12.0m)300J

6.4: a) The friction force to be overcome is

,N5.73)s/m80.9)(

kg0.30)(

25.0

b) From Eq (6.1), Fs(73.5N)(4.5m)331J The work is positive, since the worker

is pushing in the same direction as the crate’s motion

c) Since f and s are oppositely directed, Eq (6.2) gives

J

331)

m5.4)(

N5.73

6.5: a) See Exercise 5.37 The needed force is

N,2.9930

sin)25.0(30cos

)s/m80.9)(

kg30)(

25.0(sincos

keeping extra figures b) Fscos (99.2N)(4.50m)cos30386.5J, again keeping an extra figure c) The normal force is mgFsin, and so the work done by friction is

J5.386)

30sin)N2.99()s/m80.9)(

kg30)((

force and gravity act perpendicular to the direction of motion, so neither force does work e) The net work done is zero

6.6: From Eq (6.2),

J

1022.50.15cos)m300)(

N180(

Fs

6.7: 2Fscos 2(1.80106 N)(0.75103m)cos142.62109 J, or 2.6109 J to two places

6.8: The work you do is:

))m0.3(ˆm0.9((

))N40(ˆN30

s

(30N)(9.0m)(40N)(3.0m) 270Nm120Nm150J

6.9: a) (i) Tension force is always perpendicular to the displacement and does no work.

(ii) Work done by gravity is mg(y2 y1) When y1  y2, W mg 0

b) (i) Tension does no work

(ii) Let l be the length of the string W mg mg(y2 y1)mg(2l)25.1J

The displacement is upward and the gravity force is downward, so it does negative work

6.10: a) From Eq (6.6),

J

1054.1h

/km

s/m6.3

1)km/h0.50()kg1600(2

b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the

speed of any object increases the kinetic energy by a factor of four

6.11: For the T-Rex, (7000kg)((4km/hr) )2 4.32 103J

km/hr 6 3 s / m 1 2



velocity would be v 2(4.32103J)/70kg 11.1m/s, or about 40 km/h

6.12: (a) Estimate: v1m s (walking)

v2 m s (running)

m70 kg

Walking: (70kg)(1m/s)2 35J

2 1 2 2

K

Work is done by gravity and friction, so Wtot W mg W f

W mg mg(y2  y1)mgh

W f fs(kmgcos)(h/sin)kmgh/tan

Substituting these expressions into the work-energy theorem and solving for v gives 0

v0  2gh(1k /tan)

6.14: (a)

)m0.15)(

s/m80.9(2)s/m0.25(22

12

1

2 2

2 f 0

2 0

2 f

v

mv mv

mgh

KE W

s/m3.30

(b)

)s/ms80.9(2

0)s/m3.30(g

2

12

1

2

2 2 2

f

2 0

2 0

2 f

mv mv

mgh

KE W

mgh mv and v 2gh, same as if had been dropped

from height h The work done by gravity depends only on the vertical displacement of the

object

When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller

c) h15.0m, so v 2gh 17.1s

6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the

work done by friction, by a factor of four With the stopping force given as being

independent of speed, the distance must also increase by a factor of four

6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so

J

74.2m/s)

kg)(32.0145

.0)(

2/1()

2/1

W

6.18: As the example explains, the boats have the same kinetic energy K at the finish line,

so (1/2)m A v2A (1/2)m B v2B, or, with m B 2m A,v2A 2v B2 a) Solving for the ratio of the speeds, v A/v B  2 b) The boats are said to start from rest, so the elapsed time is the distance divided by the average speed The ratio of the average speeds is the same as the ratio of the final speeds, so the ratio of the elapsed times is t B /t A v A /v B  2

6.19: a) From Eq (6.5), K2 K1/16, and from Eq (6.6), W (15/16)K1 b) No; kinetic energies depend on the magnitudes of velocities only

6.20: From Equations (6.1), (6.5) and (6.6), and solving for F,

N

0.32)

m50.2(

))s/m00.4()s/m00.6)((

kg00.8()

2 1 2 1

2 2 2

K

F

)N0.40(

))s/m00.2()s/m00.6)((

kg420.0

6.22: a) If there is no work done by friction, the final kinetic energy is the work done by

the applied force, and solving for the speed,

.s/m48.4)

kg30.4(

)m20.1)(

N0.36(22



m

Fs m

W v

b) The net work is Fs fks(Fkmg)s, so

)kg30.4(

)m20.1))(

s/m80.9)(

kg30.4)(

30.0(N0.36(2

)(

3.61m/s

(Note that even though the coefficient of friction is known to only two places, the

difference of the forces is still known to three places.)

6.23: a) On the way up, gravity is opposed to the direction of motion, and so

J4.28)m0.20)(

s/m80.9)(

kg145.0

)J4.28(2)s/m0.25(

2 1

m

W v

c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up On the way down, gravity will have done both negative and positive work on the ball, but the net work will be the same

6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq (6.1) gives

J

1176)

m0.25)(

s/m80.9)(

kg80.4





Fs mgs W

kg00.7(

)m0.3)(

N0.10(2)s/m00.4(

2 1

m

W v

v

Keeping extra figures in the intermediate calculations, the acceleration is

.s/m429.1)kg00.7/(

)s/m

s/m429.1(2)s/m00.4(

2 1

2

v

giving the same result

6.26: The normal force does no work The work-energy theorem, along with Eq (6.5),

gives

,sin22

2

m

W m

s/m80.9(



v

6.27: a) The friction force is kmg, which is directed against the car’s motion, so the net work done is kmgs The change in kinetic energy is 2

0

1 (1/2)mv K

 , which in this case is about 0.279.)

6.28: The intermediate calculation of the spring constant may be avoided by using Eq

(6.9) to see that the work is proportional to the square of the extension; the work needed

to compress the spring 4.00 cm is (12.0J)34..0000cmcm2 21.3J

6.29: a) The magnitude of the force is proportional to the magnitude of the extension or

compression;

N.64)m050.0/m020.0)(

N160(,N48)m050.0/m015.0)(

N160

b) There are many equivalent ways to do the necessary algebra One way is to note that to stretch the spring the original 0.050 m requires (0.050m) 4J

m0.050

so that stretching 0.015 m requires (4J)(0.015/0.050)2 0.360J and compressing 0.020

m requires (4J)(0.020/0.050)2 0.64J Another is to find the spring constant

m/N1020.3)m050.0()

6.30: The work can be found by finding the area under the graph, being careful of the

sign of the force The area under each triangle is 1/2 base height

)J40)(

6.32: The work you do with your changing force is

xdx dx

dx x F

m

N0.3)

N0.20()

0

9 6 0

9 6

0 2 9

6

m

N0.3(

|)N0.20

 138Nm71.4Nm209.4J or 209J

The work is negative because the cow continues to advance as you vainly attempt to push her backward

2

12

1

0

2 0

6.34: a) The average force is (80.0J)/(0.200m)400N, and the force needed to hold the platform in place is twice this, or 800 N b) From Eq (6.9), doubling the distance quadruples the work so an extra 240 J of work must be done The maximum force is quadrupled, 1600 N

Both parts may of course be done by solving for the spring constant

N/m10004m)2000(J)

6.35: a) The static friction force would need to be equal in magnitude to the spring force,

kd

mg

s or μs ((020.100.0Nkg/)(m9)(.800.086m/sm2)) 1.76, which is quite large (Keeping extra figures in

the intermediate calculation for d gives a different answer.) b) In Example 6.6, the

relation

2 1

2 k

2

12

1

mv kd

60.0(2)60.0((

)m/N0.20(

)s/m80.9)(

kg10.0(

)2(

22

2 2

2

k s

2 s 2

s k

2 s

k 2

2 1

k

mg g

k

mg m

k

d gd d

m

k v

from which v1 0.67m/s

6.36: a) The spring is pushing on the block in its direction of motion, so the work is

positive, and equal to the work done in compressing the spring From either Eq (6.9) or

Eq (6.10), (200N/m)(0.025m)2 0.06J

2 1 2 2

)J06.0(2



m

W v

6.37: The work done in any interval is the area under the curve, easily calculated when

the areas are unions of triangles and rectangles a) The area under the trapezoid is

J0

6.38: a) K 4.0J, so v 2K m  2(4.0J) (2.0kg)2.00m/s b) No work is done between x3.0m and x4.0m, so the speed is the same, 2.00 m/s c)

)m/N4000()m375.0(

)kg70(

m/N4000()

6.40: a) From Eq (6.14), with dlRd,

.sin2cos

2

0

0 2

2cot2)cos1(

sin2)cos1(

sin

0

0 0

6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the

positive work done by the spring, (1/2)kx , must be the opposite of the negative work 2

done by gravity, mgLsin θ, or

cm.7.5)

m/N640(

0.40sin)m80.1)(

s/m80.9)(

kg0900.0(2sin

At this point the glider is no longer in contact with the spring b) The intermediate calculation of the initial compression can be avoided by considering that between the point 0.80 m from the launch to the maximum distance, gravity does a negative amount

of work given by (0.0900kg)(9.80m/s2)(1.80m0.80m)sin40.00.567J, and so the kinetic energy of the glider at this point is 0.567 J At this point the glider is no longer

in contact with the spring

6.42: The initial and final kinetic energies of the brick are both zero, so the net work done

on the brick by the spring and gravity is zero, so (1 2)kd2  mgh0, or

m

53.0)m/N450/(

)m6.3)(

s/m80.9)(

kg80.1(2/

provide an upward force while the spring and the brick are in contact When this force goes to zero, the spring is at its uncompressed length

6.43: Energy(power)(time)(100W)(3600s)3.6105J

kg

70for

s1002

so2

vv0 at

08.00m/s(1.96m/s2)t

t4.08s

t

mv t

KE P

2 2

)s/m00.8)(

kg0.20

yr)/s1016.3(

)yr/J100.1

W102.3

W102

2 3

6.47: The power is PFv F is the weight, mg, so

kW

15.17s)m(2.5)sm(9.8

23% of the engine power is used in climbing

6.48: a) The number per minute would be the average power divided by the work (mgh)

required to lift one box,

s,41.1m)(0.90)sm(9.80kg)30(

hp)W(746hp)50.0(

or 84.6 min b) Similarly,

s,378.0m)(0.90)sm(9.80kg)(30

W)100(

or 22.7 min

6.49: The total mass that can be raised is

kg,2436m)

(20.0)sm(9.80

s)hp)(16.0W

(746hp)0.40(

so the maximum number of passengers is 183665.0kgkg 28

6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19),

hp

3.57 W

1066.2s)

(4.00

m)(2.80N)3800

s))m6.3(h)km1(h)km65(

hp)Whp)(746(280,000

)70.0()

70

6.52: Here, Eq (6.19) is the most direct Gravity is doing negative work, so the rope

must do positive work to lift the skiers The force F

is gravity, and F Nmg, where N

is the number of skiers on the rope The power is then

W

1096.2

)0.15(90.0coshkm6.3

sm1h)km(12.0)sm(9.80kg)(70)50(

cos)()(

6.53: a) In terms of the acceleration a and the time t since the force was applied, the

speed is vat and the force is ma, so the power is PFv(ma (at)ma2t b) The power at a given time is proportional to the square of the acceleration, tripling the acceleration would mean increasing the power by a factor of nine c) If the magnitude of the net force is the same, the acceleration will be the same, and the needed power is proportional to the time At t15.0s, the needed power is three times that at 5.0 s, or

mav mva dt

dv mv

mv dt

d dt dK

t1.00s, 2 and s0.010m Pav 0.20 W

Let M  totalmass and T  timefor onerevolution

2 2 2 3

2 2 0

2 2 2 0

2 2

3

23

42

1

42

1

22

12

)(21

T ML π

L T

π L M

dx x T

π L M

T

πx dx L

M KE

T

πx v

dx L

M dm

v dm KE

L L

53(m)(2.00kg)0.12(3

 π

KE

6.57: a) (140N)(3.80m)532J b) (20.0kg)(9.80m s2)(3.80m)(sin25)315J c) The normal force does no work

d)

J20325

cosm)(3.80)sm(9.80kg)(20.0)30.0(

cos

2 k

k k

v

6.58: The work per unit mass is (W m)gh.

a) The man does work, (9.8N kg)(0.4m)3.92J kg

b) (3.92J kg) (70J kg)1005.6%

c) The child does work, (9.8N kg)(0.2m)1.96J kg

2.8%

100kg)J70(kg)

6.59: a) Moving a distance L along the ramp, sin L, sout Lsinα, so sin1

α

IMA b) If AMAIMA,(Fout Fin)(sin sout) and so (Fout)(sout)(Fin)(sin), or Wout Win c)

d)

.)

)(

(

))(

(

out in

in out in

in

out out in

out

IMA

AMA s

s

F F s

F

s F W

W

m)(18.0)sm(9.80

J)1035.7(

w

m

m0.18

J1025.8

N408N

6.61: a)

J

1059.2)mins(60min)1.90(

m)1066.6(2kg)400,86(2

12

2

12

b) (1 2)mv2 (12)(86,400kg)((1.00m) (3.00s))2 4.80103J

6.62: a)

J3.22m)(1.5012.0cos)sm(9.80kg)(5.00)31.0(

cos

2 k

(keeping an extra figure) b) (5.00kg)(9.80m s2)sin 12.0(1.50m)15.3 J

c) The normal force does no work d) 15.3J22.3J7 J

6.63: See Problem 6.62: The work done is negative, and is proportional to the distance s

that the package slides along the ramp, W mg(sinθμkcosθ)s Setting this equal to

the (negative) change in kinetic energy and solving for s gives

)cos(sin

2)cos(sin

)2/1(

k

2 1 k

2 1

θ μ θ g

v θ

μ θ mg

mv s

m/s)2.2(

.111

1 2 2

x x

x x

x

dx k dx F W

The force is given to be attractive, soF x0 , and k must be positive If

1 2

1 1 1

and W 0 b) Taking “slowly” to be constant speed, the net force on the object is zero,

so the force applied by the hand is opposite F , and the work done is negative of that x

found in part (a), or 11 12

x x

k  , which is positive if x2  c) The answers have the same x1

magnitude but opposite signs; this is to be expected, in that the net work done is zero

E/ )(R r mg

E 2

2

r

E E

mgR r

mgR dr

mgR Fds

1

K  so v2  2K2/m 11,000m/s

6.66: Let x be the distance past P.

m04000m5.12/500

2

2 i

2 f f

2 i 0

2 i k

i f f

)s/m50.4(2

12m)/0400.0()100.0()s/m80.9(

2

12)

100.0(

2

1)1000(

2

10:

v

x A x g

v dx Ax

g

mv mgdx

μ

KE KE W KE W

100.0(2

s)/m50.4(2

2 k

)

(4

4 1

4 2 3

x

With x1 x a 1.00m and x2 x b 2.00m, W 15.0J, this work is negative

6.68: From Eq (6.7), with x1 0,

4 2 3 3

2 2 2

2

4 2

3 2

2 2 3

2

)m/N3000()m/N233()m/N0.50(

432)(

x x

x

x

c x

b x

k dx cx bx kx Fdx

J441.0))s/m70.0()s/m80.2)((

kg120.0)(

2/1()(

)2

that in this case, the tension cannot be perpendicular to the block’s velocity at all times; the cord is in the radial direction, and for the radius to change, the block must have some non-zero component of velocity in the radial direction

6.70: a) This is similar to Problem 6.64, but here  0 (the force is repulsive), and

))m1025.1()m200.0((

mN1012.2(11

17

1 9 1

2 26 2

J)1065.2(2s)/m1000.3(

27

17 2

5 2

kg1067.1(

)mN1012.2(2

2 5 27

2 26 2

1 1 2

c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00105m/s

6.71: The velocity and acceleration as functions of time are

t t

a t t dt

dx t

v( ) 2 3 2, ( )2 6a) v(t4.00s)2(0.20m/s2)(4.00s)3(0.02m/s3)(4.00s)2 2.56m/s

b) ma(6.00kg)(2(0.20m/s2)6(0.02m/s3)(4.00s)5.28N

c) (1/2)(6.00kg)(256m/s)2 19.7J

2 1

.31cos)m/N00.5(

coscos

m 50 1 m 00 1 2 2

2 1

2 1

dx F

dl F

)J39.3(2)s/m00.4(

2 1

m

W v v

6.73: a) (1/2) ( 2)

1

2 2 1

J

910)

)s/m00.5()s/m50.1)((

kg0.80)(

2/1

 b) The work done by gravity is mgh (80 0 kg)(9 80 m s )(5 20 m)   2     4 08 10 J3 ,

so the work done by the rider is 910J ( 4 08 10 J) 3 17 10 J    3    3

)1())

b x

n

b dx

x

b W

Note that for this part, for n1,x1n 0 as x b) When 0 n1, the improper integral must be used,

,)(

)1(

0

1 2

n

b W

and because the exponent on the 1

1kx  mv , and solving for the spring constant,

m

/N1003.1)

m070.0(

)s/m65.0)(

kg1200

2

2 2

2



x mv k