SOLUTION – PHY430 – EXERCISES – MARCH – JUNE 2013-06-27
Question #21 p16
21. One cubic meter (1.00 m3) of aluminium has a mass of 2.70  103 kg, and the
same volume of iron has a mass of 7.86  103 kg. Find the radius of a solid
aluminium sphere that will balance a solid iron sphere of radius 2.00 cm on an equalarm balance.
SOLUTION
When the aluminium sphere is balanced by the iron sphere, their masses should be equal.
𝑚𝑎𝑙 = 𝑚𝐹𝑒
𝜌𝑎𝑙 𝑉𝑎𝑙 = 𝜌𝐹𝑒 𝑉𝐹𝑒
4 3
4 3
𝜌𝑎𝑙 𝜋𝑟𝑎𝑙
= 𝜌𝐹𝑒 𝜋𝑟𝐹𝑒
3
3
𝑟𝑎𝑙 =
𝜌
𝑟𝐹𝑒 √ 𝜌𝐹𝑒
𝑎𝑙
3
3
𝑘𝑔
7.86 ×103 3
𝑚
= (2.00 𝑐𝑚) √
𝑘𝑔
2.70 ×103 3
𝑚
= 𝟐. 𝟖𝟔 𝒄𝒎
Question #9 p68 – hard copy text book [#13 p68 – soft copy version]
13. A roller-coaster car moves 200 ft horizontally and then rises 135 ft at an angle of 30.0°
above the horizontal. It next travels 135 ft at an angle of 40.0° downward. What is its
displacement from its starting point?
SOLUTION
𝒓3𝑥
𝒓2𝑦
𝒓2
𝒓1
30
𝒓2𝑥
𝒓3𝑦
40
𝒓3
𝒓1 = 200 𝑓𝑡
𝒓2 = 135 𝑓𝑡
𝒓3 = 135 𝑓𝑡
+𝑦
Component method
+𝑥
x-component: 𝑅𝑥 = 𝑟1𝑥 + 𝑟2𝑥 + 𝑟3𝑥
= 𝑟1 + 𝑟2 cos 30° + 𝑟3 cos 40°
= 200 + 135 cos 30° + 135 cos 40°
= 420 𝑓𝑡
1
y-component: 𝑅𝑦 = 𝑟1𝑦 + 𝑟2𝑦 + 𝑟3𝑦
= 0 + 𝑟2 sin 30° + (−𝑟3 cos 40°)
= 0 + 135 sin 30° − 135 sin 40°
= −19.3 𝑓𝑡
Angle 
Resultant displacement
from the starting point R:
𝑅𝑥 = 420 ft
𝑅
𝑅𝑦 = −19.3 𝑓𝑡
𝑅 = √𝑅𝑥2 + 𝑅𝑦2 = √4202 + (−19.3)2 = 𝟒𝟐𝟎 𝒇𝒕
tan 𝜃 =
19.3
420
19.3
⟹ 𝜃 = 𝑡𝑎𝑛−1 ( 420 ) = 𝟐. 𝟔𝟑𝒐 𝒃𝒆𝒍𝒐𝒘 𝒕𝒉𝒆 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍
Question #38 p51 – hard copy text book [#40 p51 – soft copy version]
40. A baseball is hit so that it travels straight upward after being struck by the bat. A
fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) the
ball’s initial velocity and (b) the height it reaches.
SOLUTION
𝑣 = 0 at maximum height
𝑣𝑂
(a) 𝑣 = 𝑣𝑜 − 𝑔𝑡
at maximum height the final velocity 𝑣 = 0
0 = 𝑣𝑜 − 𝑔𝑡
𝑣𝑜 = 𝑔𝑡 = (9.80)(3.00) = 𝟐𝟗. 𝟒 𝒎/𝒔
(b) 𝑣 2 = 𝑣𝑜 2 − 2𝑔(𝑦 − 𝑦𝑜 )
at maximum height 𝑣 = 0 and (𝑦 − 𝑦𝑜 ) becomes maximum
0 = 𝑣𝑜 2 − 2𝑔(𝑦 − 𝑦𝑜 )𝑚𝑎𝑥
(𝑦 − 𝑦𝑜 )𝑚𝑎𝑥 =
𝑣𝑜2
2𝑔
=
29.42
2(9.80)
= 𝟒𝟒. 𝟏 𝒎
2
Question #16 p96 – hard copy text book [#14 p96 – soft copy version]
14. A rock is thrown upward from level ground in such a way that the maximum
height of its flight is equal to its horizontal range R.
(a) At what angle  is the rock thrown?
(b) In terms of its original range R, what is the range Rmax the rock can attain if it is
launched at the same speed but at the optimal angle for maximum range?
(c) What If? Would your answer to part (a) be different if the rock is thrown with the
same speed on a different planet? Explain.
SOLUTION
𝑣𝑜𝑦
(𝑦 − 𝑦𝑜 )𝑚𝑎𝑥
𝑣𝑜

𝑣𝑜𝑥
Range = R
(a)
𝑅 = (𝑦 − 𝑦𝑜 )𝑚𝑎𝑥 ...............(1)
𝑣𝑜𝑦 = 𝑣𝑜 sin 𝜃
𝑣𝑜𝑥 = 𝑣𝑜 cos 𝜃
Determine time to reach the maximum height 𝑡𝑚𝑎𝑥 in order to calculate R:
𝑣𝑦 = 𝑣𝑜𝑦 − 𝑔𝑡
0 = 𝑣𝑜𝑦 − 𝑔𝑡𝑚𝑎𝑥
𝑡𝑚𝑎𝑥 =
𝑣𝑜𝑦
𝑔
=
𝑅 = 𝑣𝑜𝑥 𝑡
𝑣𝑜 sin 𝜃
𝑔
.......(a)
but 𝑡 = 2𝑡𝑚𝑎𝑥 [time 𝑡 to travel the whole horizontal range 𝑅 is two times the time to
reach the maximum vertical height 𝑡𝑚𝑎𝑥 ]
𝑅 = 𝑣𝑜𝑥 2𝑡𝑚𝑎𝑥 = 𝑣𝑜 cos 𝜃 2 𝑡𝑚𝑎𝑥
....(b)
𝑣𝑜 sin 𝜃
)
𝑔
Substitute (a) into (b): 𝑅 = (𝑣𝑜 cos 𝜃) (2) (
𝑣𝑦2 = 𝑣𝑜𝑦 2 − 2𝑔(𝑦 − 𝑦𝑜 )
=
2𝑣𝑜2 cos 𝜃 sin 𝜃
𝑔
...(2)
at maximum height (𝑦 − 𝑦𝑜 )𝑚𝑎𝑥 𝑣𝑦 = 0
0 = 𝑣𝑜𝑦 2 − 2𝑔(𝑦 − 𝑦𝑜 )𝑚𝑎𝑥
(𝑦 − 𝑦𝑜 )𝑚𝑎𝑥 =
2
𝑣𝑜𝑦
2𝑔
=
𝑣𝑜 2 sin2 𝜃
2𝑔
....(3)
Substitute equation (2) and (3) into equation (1):
𝑅 = (𝑦 − 𝑦𝑜 )𝑚𝑎𝑥
2𝑣𝑜2 cos 𝜃 sin 𝜃
𝑔
=
4 =
𝑣𝑜 2 sin2 𝜃
2𝑔
sin 𝜃
⟹
cos 𝜃
tan 𝜃 = 4 ⟹ 𝜃 = 𝑡𝑎𝑛−1 4 = 𝟕𝟔°
3
(b) The optimal angle for maximum range is 𝜃 = 45𝑜
2𝑣𝑜2 cos 𝜃 sin 𝜃
2𝑣𝑜2 cos 76𝑜 sin 76𝑜
=
𝑔
𝑔
𝑣𝑜2
𝑅
=
= 2.13 𝑅 ...(1)
𝑔
0.469
𝑎𝑡 𝜃 = 76°: 𝑅 =
⟹
𝑎𝑡 𝜃 = 45°: 𝑅𝑚𝑎𝑥 =
2𝑣𝑜2 cos 45𝑜 sin 45𝑜
𝑔
=
𝑣𝑜2
𝑔
= 0.469
𝑣𝑜2
𝑔
...(2)
Substitute (1) into (2): 𝑹𝒎𝒂𝒙 = 𝟐. 𝟏𝟑 𝑹
(c) The angle  would be the same since tan 𝜃 = 4 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 and it does not depend on any
other variables.
Question #44 p134 – hard copy text book [#44 p134 – soft copy version]
44. A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling
on a strap at an angle  above the horizontal (Fig.P5.44). She pulls on the strap with
a 35.0-N force, and the friction force on the suitcase is 20.0 N.
(a) Draw a free-body diagram of the suitcase.
(b) What angle does the strap make with the horizontal?
(c) What is the magnitude of the normal force that the ground exerts on the suitcase?
Figure P5.44
SOLUTION
Fp = 35.0 N
N
(a)

f
W = mg
(b)
Σ𝐹𝑥 = 𝑚𝑎𝑥
𝐹𝑝 cos 𝜃 − 𝑓 = 𝑚𝑎𝑥
𝐹𝑝 cos 𝜃 − 𝑓 = 0
cos 𝜃 =
but 𝑎𝑥 = 0 since the suitcase was pulled at a constant speed
𝑓
𝐹𝑝
20.0
𝜃 = 𝑐𝑜𝑠 −1 (35.0) = 𝟓𝟓. 𝟐°
4
(c)
Σ𝐹𝑦 = 𝑚𝑎𝑦
but 𝑎𝑦 = 0 since the suitcase was moving only in horizontal direction and
no vertical motion.
𝑁 + 𝐹𝑝 sin 𝜃 − 𝑊 = 0
𝑁 = 𝑊 − 𝐹𝑝 sin 𝜃
= 𝑚𝑔 − 𝐹𝑝 sin 𝜃
= (20.0)(9.80) − 35.0 sin 55.2
= 167 𝑁
Question #40 p161 – hard copy text book [#40 p161 – soft copy version]
40 Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed
a speed bump (called the “Holly hump”) and had it installed. Suppose a 1 800-kg car passes over a hump in a roadway that
follows the arc of a circle of radius 20.4 m as shown in Figure P6.40.
(a) If the car travels at 30.0 km/h, what force does the road exert on the car as the car passes the highest point of the hump?
(b) What If? What is the maximum speed the car can have without losing contact with the road as it passes this highest
point?
N
SOLUTION
v
R
𝑣 = (30.0
W = mg
𝑘𝑚 1000𝑚
1ℎ
)(
)(
) = 8.3 𝑚/𝑠
ℎ
1 𝑘𝑚
3600 𝑠
(a) 𝐹𝑐 =
𝑚𝑣 2
𝑅
Choose upward direction as positive and downward direction as negative.
𝑁−𝑊 =−
𝑁 =𝑊
𝑚𝑣 2
𝑅
𝑚𝑣 2
−
𝑅
= 𝑚𝑔 −
𝑚𝑣 2
𝑅
= 𝑚 (𝑔 −
𝑣2
)
𝑅
= (1800) (9.80 −
8.32
)
20.4
= 𝟏. 𝟐 × 𝟏𝟎𝟒 𝑵
5
(b) At the maximum speed, the car just begin to lose contact with the road when N = 0.
𝑁 = 𝑚𝑔 −
0 = 𝑚𝑔 −
𝑚𝑣 2
𝑅
𝑣max 𝑤ℎ𝑒𝑛 𝑁 = 0
2
𝑚𝑣𝑚𝑎𝑥
𝑅
𝑣𝑚𝑎𝑥 = √𝑔𝑅 = √(9.80)(20.4) = 𝟏𝟒. 𝟏 𝒎/𝒔
Question #41 p161 – hard copy text book [#41 p161 – soft copy version]
41. A car of mass m passes over a hump in a road that follows the arc of a circle of radius R as shown in Figure P6.40.
(a) If the car travels at a speed v, what force does the road exert on the car as the car passes the highest point of the hump?
(b) What If? What is the maximum speed the car can have without losing contact with the road as it passes
this highest point?
N
SOLUTION
v
R
(a) 𝐹𝑐 =
W = mg
𝑚𝑣 2
𝑅
Choose upward direction as positive and downward direction as negative.
𝑁−𝑊 =−
𝑁 =𝑊
𝑚𝑣 2
𝑅
𝑚𝑣 2
− 𝑅
= 𝑚𝑔 −
𝑚𝑣 2
𝑅
= 𝑚 (𝑔 −
𝑣2
)
𝑅
(b) At the maximum speed, the car just begin to lose contact with the road when N = 0.
𝑁 = 𝑚𝑔 −
0 = 𝑚𝑔 −
𝑚𝑣 2
𝑅
𝑣max 𝑤ℎ𝑒𝑛 𝑁 = 0
2
𝑚𝑣𝑚𝑎𝑥
𝑅
𝑣𝑚𝑎𝑥 = √𝑔𝑅
6
Question #14 p193 – hard copy text book [#14 p193– soft copy version]
14. The force acting on a particle varies as shown in Figure P7.14. Find the work done by the force on the particle as it
moves
(a) from x = 0 to x = 8.00 m,
(b) from x = 8.00 m to x = 10.0 m, and
(c) from x = 0 to x = 10.0 m.
SOLUTION
(a) 𝑊0−8 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ
1
= 2 (8)(6)
= 24 𝐽
1
(b) 𝑊8−10 = − 2 (3)(4) = −6 𝐽
(c) 𝑊0−10 = 𝑊0−8 + 𝑊8−10
= 24 − 6
= 18 𝐽
Question #35 p195 – hard copy text book [#35 p195– soft copy version]
35. A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile
driver falls 5.00 m before coming into contact with the top of the beam, and it drives
the beam 12.0 cm farther into the ground before coming to rest. Using energy
considerations, calculate the average force the beam exerts on the pile driver while
the pile driver is brought to rest.
SOLUTION
Gravitational potential energy lost by the pile driver = Work done on the beam
𝑚𝑔ℎ = 𝐹𝑎𝑣𝑒 𝑑
𝐹𝑎𝑣𝑒 =
=
𝑚𝑔ℎ
𝑑
(2100)(9.80)(5.00+0.120)
0.120
= 𝟖. 𝟕𝟖 × 𝟏𝟎𝟓 𝑵 𝒖𝒑𝒘𝒂𝒓𝒅
7
Question #55 p273 – hard copy text book [#55 p273– soft copy version]
55. A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of u 5 60.08 with the surface.
It bounces off with the same speed and angle (Fig. P9.55). If the ball is in contact with the wall
for 0.200 s, what is the average force exerted by the wall on the ball?
SOLUTION
𝑃𝑓𝑦 𝑃𝑖𝑦
𝑃𝑓 𝜃
𝜃
𝑃𝑓𝑥
𝑃𝑖
𝑃𝑖𝑥
x
𝑃𝑖 𝜃
𝑃𝑖 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙
= 𝑚𝑣
𝑃𝑓 = 𝑓𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙
= 𝑚𝑣
𝐹𝑎𝑣𝑒 =
∆𝑃
∆𝑡
x-component:
𝐹𝑎𝑣𝑒−𝑥 =
∆𝑃𝑥
∆𝑡
=
𝑃𝑓𝑥 −𝑃𝑖𝑥
∆𝑡
=
−𝑃𝑓 sin 𝜃−𝑃𝑖 sin 𝜃
∆𝑡
=
−𝑚𝑣 sin 𝜃−𝑚𝑣 sin 𝜃
∆𝑡
=
−2𝑚𝑣 sin 𝜃
∆𝑡
=
−2(3.00)(10.0)
0.200
= −260 𝑁
y-component:
∆𝑃𝑦
𝑃𝑓𝑦 − 𝑃𝑖𝑦
𝑃𝑓 cos 𝜃 − 𝑃𝑖 cos 𝜃
𝑚𝑣 cos 𝜃 − 𝑚𝑣 cos 𝜃
𝐹𝑎𝑣𝑒−𝑦 =
=
=
=
=0
∆𝑡
∆𝑡
∆𝑡
∆𝑡
 𝐹𝑎𝑣𝑒 = 𝐹𝑎𝑣𝑒−𝑥 = −𝟐𝟔𝟎 𝑵 to the left
8
Question #35 p311 – hard copy text book [#35 p311– soft copy version]
35. Find the net torque on the wheel in Figure P10.35 about the axle through O,
taking a = 10.0 cm and b = 25.0 cm.
SOLUTION
r1
r3
r3
F3 = 12.0 N
F1 = 10.0 N
r1
r2
r2
F2 = 9.00 N
𝝉𝑛𝑒𝑡= 𝝉1 + 𝝉2 + 𝝉3
= 𝒓1 × 𝑭1 + 𝒓2 × 𝑭2 + 𝒓3 × 𝑭3
= (0.250)(10.0) sin 90° + (0.250)(9.0) sin 90° − (0.100)(12.0) sin 90°
= 𝟑. 𝟓𝟓 𝑵𝒎 into the page
Question #31 p459 – hard copy text book [#33 p459 – soft copy version]
33. A simple pendulum is 5.00 m long. What is the period of small oscillations for this
pendulum if it is located in an elevator
(a) accelerating upward at 5.00 m/s2?
(b) Accelerating downward at 5.00 m/s2?
(c) What is the period of this pendulum if it is placed in a truck that is accelerating
horizontally at 5.00 m/s2?
SOLUTION
𝐿
a) 𝑇 = 2𝜋√𝑔′
𝑔′ = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑒𝑛𝑑𝑢𝑙𝑢𝑚
When the pendulum is in an elevator accelerating upward,
𝑔′ = 𝑔 + 𝑎
𝐿
5.00
𝑇 = 2𝜋√𝑔+𝑎 = 2𝜋√9.80+5.00 = 𝟑. 𝟔𝟓 𝒔
b) When the pendulum is in an elevator accelerating downward,
𝑔′ = 𝑔 − 𝑎
𝐿
5.00
𝑇 = 2𝜋√𝑔−𝑎 = 2𝜋√9.80−5.00 = 𝟔. 𝟒𝟏 𝒔
9
a) When the pendulum is inside a truck horizontally,
𝒂
𝑔′ = √𝑔2 + 𝑎2 = √9.802 + 5.002 = 11.0 𝑚/𝑠 2
g’
𝐿
𝒈
5.00
𝑇 = 2𝜋√𝑔′ = 2𝜋 √11.0 = 𝟒. 𝟐𝟒 𝒔
Question #19 p425 – hard copy text book [#19 p425 – soft copy version]
19. A backyard swimming pool with a circular base of diameter 6.00 m is filled to
depth 1.50 m.
(a) Find the absolute pressure at the bottom of the pool.
(b) Two persons with combined mass 150 kg enter the pool and float quietly
there. No water overflows. Find the pressure increase at the bottom of the pool
after they enter the pool and float.
SOLUTION
h = 1.50 m
d = 6.00 m
a) 𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚 + ℎ𝜌𝑔
= 1.01 × 105 + (1.50)(1000)(9.80)
= 1.16 × 105 𝑃𝑎
b) 𝑃𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 =
𝐹
𝐴
=
𝑊
𝐴
=
𝑚𝑔
𝜋𝑟 2
=
(150)(9.80)
𝜋3.002
= 𝟓𝟐 𝑷𝒂
Question #12 p593 – hard copy text book [#8 p593 – soft copy version]
8. An aluminum cup of mass 200 g contains 800 g of water in thermal equilibrium at
80.0°C. The combination of cup and water is cooled uniformly so that the
temperature decreases by 1.50°C per minute. At what rate is energy being removed
by heat? Express your answer in watts.
10
SOLUTION
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑢𝑝 𝑎𝑛𝑑 𝑤𝑎𝑡𝑒𝑟 𝑏𝑦 ℎ𝑒𝑎𝑡 =
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 =
∆𝑄
∆𝑡
∆𝑇
℃
1 𝑚𝑖𝑛
℃
= (1.50
)(
) = 0.025
∆𝑡
𝑚𝑖𝑛
60 𝑠
𝑠
The reduction in temperature for the cup and water = ∆𝑇
The heat removed from the cup and water = ∆𝑄
∆𝑄 = 𝑚𝑐𝑢𝑝 𝑐𝑐𝑢𝑝 ∆𝑇 + 𝑚𝑤𝑎𝑡𝑒𝑟 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇 ....(1)
Devide equation (1) by ∆𝑡:
∆𝑄
∆𝑇
∆𝑇
= 𝑚𝑐𝑢𝑝 𝑐𝑐𝑢𝑝
+ 𝑚𝑤𝑎𝑡𝑒𝑟 𝑐𝑤𝑎𝑡𝑒𝑟
∆𝑡
∆𝑡
∆𝑡
= (0.200)(900)(0.025) + (0.800)(4186)(0.025)
= 𝟖𝟖. 𝟐 𝑾
Question #19 p484– hard copy text book [#19 p484 – soft copy version]
19. (a) Write the expression for y as a function of x and t in SI units for a sinusoidal
wave travelling along a rope in the negative x direction with the following
characteristics: A = 8.00 cm,  = 80.0 cm, f = 3.00 Hz, and y(0, t) = 0 at t = 0.
(b) What If? Write the expression for y as a function of x and t for the wave in part
(a) assuming y(x, 0) 5 0 at the point x 5 10.0 cm.
SOLUTION
a)
General equation for a travelling wave in the negative x-direction:
𝑦 = 𝐴 sin(𝑘𝑥 + 𝜔𝑡 + 𝜙)
𝑘=
2𝜋
𝜆
=
2𝜋
0.800
= 2.5𝜋 ; 𝜔 = 2𝜋𝑓 = 2𝜋(3.00) = 6𝜋
𝑦 = 0.0800 sin(2.5𝜋𝑥 + 6𝜋𝑡 + 𝜙)
Condition: 𝑦 (0, 𝑡) = 0 𝑎𝑡 𝑡 = 0
0 = 0.0800 sin(2.5𝜋(0) + 6𝜋(0) + 𝜙)
0 = 0.0800 sin 𝜙
sin 𝜙 = 0 ⇒ 𝜙 = 0
∴ 𝑦 = 0.0800 sin(2.5𝜋𝑥 + 6𝜋𝑡 + 0)
𝒚 = 𝟎. 𝟎𝟖𝟎𝟎 𝐬𝐢𝐧(𝟐. 𝟓𝝅𝒙 + 𝟔𝝅𝒕)
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b) Condition: 𝑦 (𝑥, 0) = 0 𝑎𝑡 𝑥 = 10.0 𝑐𝑚 = 0.100 𝑚
Substitute the above condition into the following equation:
𝑦 = 0.0800 sin(2.5𝜋𝑥 + 6𝜋𝑡 + 𝜙)
We get,
0 = 0.0800 sin(2.5𝜋(0.100) + 6𝜋(0) + 𝜙)
0 = 0.0800 sin(2.5𝜋(0.100) + 𝜙)
sin(0.25𝜋 + 𝜙) = 0
0.25𝜋 + 𝜙 = 0
𝜙 = −0.25𝜋
∴ 𝒚 = 𝟎. 𝟎𝟖𝟎𝟎 𝐬𝐢𝐧(𝟐. 𝟓𝝅𝒙 + 𝟔𝝅𝒕 − 𝟎. 𝟐𝟓𝝅 )
Question #33 p595– hard copy text book [#31 p594 – soft copy version]
31. An ideal gas initially at 300 K undergoes an isobaric expansion at 2.50 kPa. If the
volume increases from 1.00 m3 to 3.00 m3 and 12.5 kJ is transferred to the gas by
heat, what are
(a) the change in its internal energy and
(b) its final temperature?
SOLUTION
a) Δ𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊 ;
𝑊 = −𝑝(𝑉𝑓 − 𝑉𝑖 ) = −(2.5 × 103 )(3.00 − 1.00) = −2.5 × 103 = −2.5𝑘𝐽
∴ Δ𝐸𝑖𝑛𝑡 = 12.5 + (−2.5) = 𝟕. 𝟓 𝒌𝑱
b) 𝑃𝑉 = 𝑛𝑅𝑇
𝑉
𝑇
=
𝑛𝑅
𝑃
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ; for isobaric process p = constant
𝑉1
𝑉2
=
𝑇1
𝑇2
𝑇2 =
𝑉2
3.00
𝑇1 = (
) (300) = 𝟗𝟎𝟎 𝑲
𝑉1
1.00
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Question #27 p594– hard copy text book [#29 p594 – soft copy version]
29. A thermodynamic system undergoes a process in which its internal energy
decreases by 500 J. Over the same time interval, 220 J of work is done on the
system. Find the energy transferred from it by heat.
SOLUTION
Δ𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊
Q = Δ𝐸𝑖𝑛𝑡 − 𝑊 = −500 − 220 = −𝟕𝟐𝟎 𝑱 the amount of energy transferred
by heat from the system
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