geas mar2014 th2 - solutions

Excel Review Center ECE Refresher Course GEAS 2

Baguio: PiRC Building, Gapuz Centrum II, 80 T. Alonso St., New Lucban CP: 0928 934 1637 Manila: CMFFI Bldg. R. Papa St. Tel. 7365291

1. If the coefficient of sliding friction…

N = W = 150 lb

F = µN = 0.05( ) 150( )F = 7.5 lb

Fforward = F = 7.5 lb → Ans

2. A 3200 lb car traveling with a speed…

Fc = m v2

Fc =3200 lb

32.2 fts2

88 fts

⎛⎝⎜

⎞⎠⎟

484 ft

Fc = 1590 lb → Ans

3. A 400 kN block is resting on a rough…

N = WN = 400 kNF = µNF = 0.40( ) 400( )F = 160 kNP = F = 160 kN → Ans

4. A skier is on a 25° slope. His mass…

N = W cos25°N = 80( ) 9.81( )cos25°

N = 711.27 NF = µNF = 0.08( ) 711.27( )F = 56.9 N

∑F = ma80( ) 9.81( )sin25° − 56.90 = 80a

a = 3.43 m/s2 → Ans

5. A trapezoid with parallel bases…

y = 2a +b

a +bh3=

2 2( ) + 42+ 4

33= 1.33→ Ans

6. A 250-g stone is attached to a 50-cm…

F = mv2

0.25 kg( ) 2 m/s( )2

0.50 m= 2N→ Ans

7. A 600 N block rests on a 30° plane…

Fx = 0∑

Pcos30° = F + 600sin30°Pcos30° = µN+ 600sin30° →Eq. 1

Fy = 0∑

N = 600cos30° +Psin30° → Eq. 2 Substitute Eq. 2 in Eq. 1:

Pcos30 = 0.2 600cos30 +Psin30⎡⎣ ⎤⎦ +

600sin300.866P = 103.923 + 0.1P + 300

P = 527.31 N → Ans

8. A pipeline crossing a river is…

w = wp + wc = 14 +1= 15 kg/m

H = wL2

8d= 15(100)2

8(2)= 9375 kg

T = wL2

⎛⎝⎜

⎞⎠⎟

T = 15(100)2

⎛⎝⎜

⎞⎠⎟

+ 93752

T = 9405 kg → Ans

9. A cable suspended from a level…

Using squared property of parabola:

30= x2

x2 = 1002

3x = 57.8 → Ans

10. Determine the moment of inertia…

I= 2mr2

I= 2(10)(3)2

3I= 60 → Ans

11. An inverted cone (apex down) has….

I= 3mr2

I= 3(10)(2)2

10I= 12 → Ans

12. A car is traveling at 60 kph….

a = ΔvΔt

=40 km

hr1000 m

1 km⎛⎝⎜

⎞⎠⎟

1 hr3600 s

⎛⎝⎜

⎞⎠⎟

6 sa = 1.85 m/s2 → Ans

13. A man, standing on a cliff overlooking…

s = vot ±12

−y = 0 − 12

9.81( ) 2( )2 ; y = 19.6 m → Ans

14. A tennis ball is dropped from the…

s = vot ±12

−40 = 0 − 12

9.81( ) t2 ; t = 2.86 s → Ans

15. Two blocks weighing 8 kg and…

Block 1:

Block 2:

Solve for a in eq. 1 and eq. 2: a = 1.4

16. A car travels at the constant speed… The average speed is equal to that of the constant speed which would be required for the object to travel the same distance d in the same time t.

25° N

v =dtotal

ttotal

v = 20 + 20 + 202030

+ 2040

+ 2050

v = 38.2 mph → Ans

17. During the takeoff, an Airbus A380…

18. A dragster reaches a quarter mile…

v2 = vo2 ± 2as

802 = 0 + 2a 402( )a = 7.96 m/s2

v = vo ± at

80 m / s = 0 + 7.96 m/s2( ) tt = 10 s → Ans

19. A ball is released from rest at a…

20. An airplane lands on a carrier deck…

21. A ball is thrown with an initial velocity…

This is a problem in projectile motion. Analyze first the horizontal component of the motion. Note: The ball is not accelerating horizontally.

vx = vox ± axtvx = 160cos53° + 0vx = 96 ft/s → Ans

Analyze the vertical component of the motion. Note: The vertical acceleration is due to gravity (ay = g)

Get the resultant of the two components:

v = vx2 + vy

v = 962 + 642

v = 115.4 ft/s → Ans

Analyzing just the vertical component of the motion, at the highest point the velocity (vertical component only) is zero:

vy = voy ± ayt

0 = 160sin53° − 32.2( ) tt = 4 s → Ans

2 sin2θg

R =1602 sin2 53( )

32.2R = 764 ft → Ans

24. A 65-lb horizontal force is sufficient…

N = W = 65 lb

Fapplied = F = 1200 lb

F = µN1200 = µ 65( )

µ = 0.054 → Ans

25. A 1000-gram mass slides down an…

s = Vot ±12

81= 0 + 12

a 0.6( )2

a = 450 cm/s2

F = maF = 1000 g( ) 450 cm/s2( )F = 450,000 dynes → Ans

26. A baseball pitcher throws a ball…

W = mg

m = Wg

32.2 fts2

m = 0.01 slugs

F = maF = 0.01 slugs( ) 480 ft/s2( )F = 5 lb → Ans

27. The outside curve on a highway…

sinθ = 430

θ = 7.66°

tanθ = v2

tan7.66° = v2

150( ) 32.2( )v = 25.4 ft/s → Ans

28. A satellite is placed in a circular…

R = Rearth +100

R = 6.4 ×106m+1.6 ×105mR = 6.5 ×106m

Fgravity = Fc

mg = mv2

g = v2

v = 9.81ms2

⎛⎝⎜

⎞⎠⎟

6.5 ×106m( )v = 8000 m

T = 2πRv

T =2π 6.5 ×106m( )

8000 ms

T = 85 min → Ans

29. A cable 800 m long weighing 15.5…

30. From the cable in the previous…

x = cloge

s + yc

x = 300loge

400 + 500300

x = 330 m → Ans

31. On a certain stretch of the railroad…

32. Water drops from a faucet at the rate…

v = vo ± at

−3 = 0 − 9.81( ) tt = 0.306 s

s = Vot ±12

−h = 0 − 12

9.81( ) 0.306( )2

h = 0.459 m → Ans

33. Suppose that you throw a ball…

v2 = vo2 ± 2as

0 = 102 − 2 9.81( )hh = 5.10 m

hmax = 5.10 + 2hmax = 7.10 m → Ans

34. A man weighing 70 kg is in an…

35. When a 3000 N boat is moving at…

v2 = vo2 ± 2as

0 = 32 − 2as

a = 4.5s

a = R = 30V

V =v1 + v2

V = 3 + 02

V = 1.5 m/s

30009.81

= 30 1.5( )s = 30.6 m→ Ans

36. A projectile is fired with an initial…

y = x tanθ − gx2

2v2 cos2 θ

−80 = x tan30° −9.81( )x2

2 60( )2cos2 30°

x = 422 m → Ans

37. A rocket is released from a jet… Vertical component of the motion:

y = Voyt ±12

−2400 = 0 − 12

9.81( ) t2

t = 22.12 s → Ans

Horizontal component of the motion:

x = Voxt ±12

x = 1200( )10003600

22.12( ) + 12

0.6( ) 9.81( ) 22.12( )2

x = 8813 m

38. An airplane makes a turn in a…

tanθ = v2

tanθ =215 m

s⎛⎝⎜

⎞⎠⎟

1750 m( ) 9.81 ms2

⎛⎝⎜

⎞⎠⎟

θ = 69.6° → Ans

39. A flywheel 6 ft in diameter accelerates…

α =ω f − ω i

α =4 rev

min× 2π rad

1 rev× 1 min

60 s⎡

⎣⎢

⎦⎥ − 0

10 sα = 0.42 rad/s2 → Ans

40. What is the velocity of a particle after…

v = x3 − 2x2 − 5x + 4

v = 5( )3 − 2 5( )2− 5 5( ) + 4

v = 54 m/s → Ans

41. What is the maximum speed at which…

Note: since the curve is “unbanked”, angle of elevation θ = 0; Φ is called angle of friction and is equal to Arctan µ where µ is the coefficient of friction

tan θ + φ( ) = vmax( )2

tan 0 + tan−10.80( ) = vmax( )2

60( ) 9.81( )vmax = 21.7 m/s → Ans

42. A man standing at the foot of a…

h = 12

32.2( ) 1( ) 2( ) = 32.2 s → Ans

43. A car starting from rest accelerates…

v = vo ± at

120 = 0 + a 4( ); a = 30 ft/s2

s = vot ±12

s = 0 + 12

30 ft/s2( ) 2( )2; s = 60 ft → Ans

44. A car is traveling at 40 kph…

v = vo ± at

0 = 40 kmhr

× 1000 m3600 s

⎛⎝⎜

⎞⎠⎟− 6 m

⎛⎝⎜

⎞⎠⎟

t = 1.85 s → Ans

45. A projectile is fired from the top…

Vertical component of the motion:

sy = voyt ±12

−200 = 1314sin45( ) t − 12

32.2( ) t2

t = 58 s

Horizontal component of the motion:

sx = voxt ±12

sx = 1314cos45( ) 58( ) + 0

sx = 53,890 ft → Ans

46. Two cars A and B are traveling in…

d = 100 m + 70( )10 − 60( )10⎡⎣ ⎤⎦d = 200 m → Ans

47. A block weighing 400 N is pulled…

∑Fhorizontal = 0Phorizontal − f = 0

220cos40° − µ 400( ) = 0

µ = 0.42 → Ans

48. A moon revolves around the earth…

aN = v2

ωr( )2

r= ω2r

2.7 mms2 × 1 m

1000 mm= 2.7 ×10−6( )2

r = 3.7 ×106 m → Ans

49. A car enters a 350-m radius curve…

vo = 50 kmhr

= 13.89 ms

v2 = vo2 + 2ats

v2 = 13.892 + 2 2( ) 320( )v = 38.38 m

an =v2

38.38 ms

⎛⎝⎜

⎞⎠⎟

320= 4.6 m

atotal = an2 + at

atotal = 4.62 + 22 = 5 ms2 → Ans

50. A tennis ball is projected upward…

Upward motion:

v = vo ± at0 = 120 − 9.81t ; t = 12.23 s

It will also take 12.23 s to move

downwards. Therefore, total time is 25 s. 51. What is the resultant force on a body…

F = maF = 48 kg( ) 6 m/s2( )F = 288 N → Ans

52. Two horses on opposite banks…

ΣFx = −200cosθ − 240cos 60 − θ( )ΣFy = 0

ΣFy = 200sinθ − 240sin 60 − θ( )0 = 200sinθ − 240sin 60 − θ( )θ = 33°

R = ΣFx2 + ΣFy

R = −381.5( )2+ 0

R = 381.5 kN → Ans

53. What is the upward acceleration…

∑Fy = ma

R− W = ma1000 − 90( ) 9.81( ) = 90a

a = 1.3 m/s2 → Ans

54. A car (mass = 1800 kg) is traveling…

F = ma = 1800 kg( ) −20 cms2

⎛⎝⎜

⎞⎠⎟

1 m100 cm

F = −360 N → Ans

55. What is the weight of a person on…

Won moon =

120( ) = 20 lb → Ans

56. Two blocks of mass 6 kg and m kg…

W = mg174 = 6 +m( ) 9.67( ); m = 12 kg → Ans

57. What is the combined kinetic energy…

KE = 12

225 kg( ) 40 kmhr

× 1000 m3600 s

⎛⎝⎜

⎞⎠⎟

KE = 13.89 kJ → Ans

58. If a certain object with mass of 15 kg…

PE = mgh

PE = 15 kg( ) 13

9.81 ms2

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥ 100 ft × 1 m

3.3 ft⎡

⎣⎢

⎦⎥

PE = 1500 J → Ans

59. If a car with initial velocity, v…

KE1 =12

mv2 = 0.5mv2

KE2 =12

m 5v( )2= 1

2m 25v2( ) = 12.5mv2

ΔKE = 12.5mv2 − 0.5mv2

ΔKE = 12mv2 = 24 12

mv2⎛⎝⎜

⎞⎠⎟

ΔKE = 24x → Ans

60. Find the average power necessary…

P = Wt= mgh

30( ) 9.81( ) 10( )60

P = 49 W → Ans

61. A charger delivers current of 6 A…

P = IV; Et= IV

E = IVt = 6 A( ) 12 V( ) 6 hr( )3600 s1 hr

E = 1.55 ×106 J → Ans

62. Find the volume (in cm3) of a piece…

ρ = mv

; v = mρ= 70 g

2.33 gcm3

= 30cm3 → Ans

63. A golf ball was hit with a velocity of…

x θ α

200 kN

240 kN

Ft = Δmv

F = Δmvt

=0.050 kg( ) 90 m

s⎛⎝⎜

⎞⎠⎟− 0

0.3 sF = 15 N → Ans

64. A toaster connected to a 220 V…

P = V2

Et= V2

220 V( )2

15 ohmsE = 194 kJ → Ans

65. A 1-A lamp on a 220 V line operates…

E = Pt = IV( ) tE = 1 A( ) 220 V( ) 10 days( )24 hr

1 day= 52.8 kW ⋅hr

cost = 52.8 kW ⋅hr( ) P2.501 kW ⋅hr

= P132→ Ans

66. A 200 gram apple is thrown from…

ΔKE = 12

mvf2 − 1

ΔKE = 12

m vf2 − vi

2( )ΔKE = 1

20.2 kg( ) 502 − 202( )

ΔKE = 210 J × 1 cal4.186 J

ΔKE = 50.17 cal → Ans

67. What is the energy of emitted light…

E = hf

E = 6.63×10−34 Js( ) 6.88×10−19s−1( )E = 4.56×10−19J → Ans

68. A body initially at rest is acted upon…

Ft = ΔmvFt = mv2 −mv1

18 5( )−12t = 0 − 0

t = 7.5 s

69. Three concurrent forces at the origin…

d = 32 + 52 + 72 = 9.11

20009.11

3Fx = 660 N → Ans

70. A meteor is moving away from a…

v =v1 + v2

1+v1v2

v = 0.75c + 0.48c

1+0.75c( ) 0.48c( )

v = 0.904c → Ans

71. A 2-m long pendulum is pulled aside…

T = 2π L

g= 2π 2

9.81= 2.8 s → Ans

72. Two electrons (Q1 and Q2) of

electrical…

F = kQ1Q2

d2 = 9 ×109( ) 0.003( ) 0.005( )32

F = 15,000 N → Ans

73. A block, weighing 100 lbs is

suspended…

;keff = 12 lbin

T = 2π mkeff

32.2 fts2

12 lbin

× 12 in1 ft

T = 0.92 s → Ans

74. How long will it take for sound to

travel…

Note: Solve first for the speed of sound in air at 20°C. The speed of sound in air at 0°C is approximately 330 m/s

v = 330 1+ 20273

= 341.9 ms

t = dv= 4000 m

341.9 ms

= 11.7 s → Ans

75. An electron at rest has a mass of…

E = mc2

E = 9.11×10−31( ) 3 ×108( )2

E = 8.198 ×10−14 J

E = mV2

8.198 ×10−14 J = 2 9.11×10−31( )v2

v = 2.12×108 m/s → Ans

76. A choir is composed of 39 singers…

I= 10log IIo−10log 39I

IoI= −16 dB → Ans

77. What is the intensity level at a point…

SIL = 10log

4π 20m( )2

10−12 Wm2

= 86 dB → Ans

78. A sound has an intensity of 10-10…

I= 10log IIo

I= 10log10−10 W

⎛⎝⎜

⎞⎠⎟

100 cm1 m

⎛⎝⎜

⎞⎠⎟

10−12 Wm2

= 60 dB

79. A human ear can distinguish a…

Note: The time it travels one way is just 0.10 s.

d = vt = 1150 ft

s⎛⎝⎜

⎞⎠⎟

0.10s( ) = 115ft → Ans

80. A ray of light travels from air at an…

n = cv

1.33 = 3 ×108

v;v = 2.26 ×108 m

s→ Ans

81. Find the period of the 100-cm…

T = 2 1 s( ) = 2 s → Ans

82. The speed of light in a particular…

83. A 10-g mass attached to a spring…

Get the spring constant first using the period of the motion given in the problem:

ω = 2π f1f= 2π

T = 2πω

= 2π mk

2 = 2 10 0.010k

k = 0.10 N/m

Use the spring formula:

F = kxF = 0.10( ) 0.10( )F = 10−2 N

84. The star nearest Earth, Alpha

Centauri…

d = 4.3 LY × 3×108 m 1 s

× 31,536,000 s 1 year

d = 4.068×1016 m

d = vt

t = dv

= 4.068×1016 m

0.95 3×108 ms

⎛⎝⎜

⎞⎠⎟

t = 142.74×106 st = 4.53 yr

85. The star nearest Earth, Alpha

Centauri…

Time based on an Earth clock

d = 4.3 LY × 3×108 m 1 s

× 31,536,000 s 1 year

d = 4.068×1016 m

d = vt

t = dv

= 4.068×1016 m

0.95 3×108 ms

⎛⎝⎜

⎞⎠⎟

t = 142.74×106 st = 4.53 yr

Use time dilation (since clock is now moving):

t = t'

1− v2

4.53 = t'

1−0.95c( )2

t' = 1.41 years

86. An observer moves past a meter…

l' = l 1− v2

l' = 1 m( ) 1−0.5c( )2

l' = 0.866 m

87. An observer measures the length…

l' = l 1− v2

0.5 m = 1 m( ) 1− v2

3×108 ms

⎛⎝⎜

⎞⎠⎟

v = 2.59×108 ms

88. A rocket is moving with a velocity…

l' = l 1− v2

l' = 5 m( ) 1−0.385c( )2

l' = 4.61 m

89. A spaceship moving away from the…

V = v1 + v2

1+v1v2

V = 0.75c + 0.75c

1+0.75c( ) 0.75c( )

V = 1.5c1+ 0.5625

V = 0.96c

90. An incident ray from water makes…

For reflection, angle of incidence is always equal to the angle of reflection. Therefore, the angle of reflection is also 55 degrees.

91. A diverging lens has a focal length of…

Use Thin-Lens Equation (it is assumed that the focal length is much greater than the thickness of the lens):

−5 cm= 1

4 cmi = −2.22 cm (virtual)

f is + for a converging lens (thicker at the center than at the sides)

f is – for a diverging lens (thinner in the center than at the sides) o is + if object is on the same side of the lens as the incident light (real object) o is – if the object is on the other side i is + if the image is on the side of the lens opposite the incident light (real) i is – if the image is on the same side (virtual)

Use formula for Lateral Magnification:

m = − io

m = −−2.2( )

4m = 0.55 (erect, smaller)

Note: + value of m means image is erect - value of m means image is inverted m = 1 means object and image have same size m < 1 means image is smaller m > 1 means image is larger

92. A penny is placed 4.0 cm in front…

15 cm= 1

4 cmi = −5.45 cm (virtual)

m = − io

m = −−5.45( )

4m = 1.4 (erect, larger)

93. A flea is located 3.0 cm from a… Spherical mirrors are curved mirrors used as image-forming devices. You can still use the thin-lens equation. Focal length of the spherical mirror is half its radius.

−5 cm= 1

3 cmi = −1.9 cm (virtual) → Ans

94. A far-sighted classmate is unable…

Nearsightedness (myopia) can be corrected by using a diverging lens while Farsightedness (hyperopia) can be corrected by using a converging lens.

D = 1f

D = 1i+ 1

D = 1−1.25

+ 10.3

D = 2.5 diopters → Ans

Note: i is – because the image and object are on the same side of the lens (image must be virtual)

95. A photographer has an 8x magnifier…

Magnifying glasses are converging lenses.

M= 25 cmf

8 = 25 cmf

f = 3.1 cm → Ans

96. A biology student wishes to use…

M= 25 cmf

M= 25 cm6 cm

M= 4.17 = 4x → Ans

97. For the biology student in the…

Mmax =25 cm

M= 25 cm6 cm

M= 5x → Ans

98. A laboratory microscope has a 20X…

A typical microscope consists of a tube with a converging lens at both ends. The lens close to the object is called the objective lens. The lens through which one looks is called the eyepiece/ocular lens

Mobjective =16

fobjective

Meyepiece =25

feyepiece

Mtotal = MobjectiveMeyepiece

Mobjective =16 cmfobjective

20 = 16 cmfobjective

fobjective = 0.8 cm → Ans

99. From the previous problem…

Meyepiece =25 cmfeyepiece

10 = 25 cmfeyepiece

feyepiece = 2.5 cm → Ans

100. Given the microscope in…

Mtotal = MobjectiveMeyepiece

Mtotal = 20( ) 10( )Mtotal = 200x

200( ) 9 nm( ) = 1.8µm → Ans

geas mar2014 th2 - solutions

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