# Fundamentos da termodinamica- exercicios resolvidos

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2-1

CHAPTER 2
The correspondence between the problem set in this fifth edition versus the
problem set in the 4'th edition text. Problems that are new are marked new and
those that are only slightly altered are marked as modified (mod).
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

Old
4 mod
new
new
7 mod
2 mod
new
new
new
5 mod
6
8 mod
new
9 mod
10 mod
11
new
new
16 mod
new
12

New
2122
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39E
40E

Old
13
14
15
17
18
new
19
20
21
22
23
24
new
25 mod
26 mod
27 mod
28
29
31E mod
32E

New
41E
42E
43E
44E
45E
46E
47E
48E
49E

Old
33E mod
34E mod
35E
36E
37E
38E
39E
40E
41E

2-2
2.1

The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665
m/s2. What is the force needed to hold a mass of 2 kg at rest in thisgravitational
field ? How much mass can a force of 1 N support ?
Solution:
ma = 0 = ∑ F = F - mg
F = mg = 2 × 9.80665 = 19.613 N
F = mg

2.2

=>

m = F/g = 1 / 9.80665 = 0.102 kg

A model car rolls down an incline with a slope so the gravitational “pull” in the
direction of motion is one third of the standard gravitational force (see Problem
2.1). If the car has a mass of 0.45 kg. Find theacceleration.
Solution:
ma = ∑ F = mg / 3
a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2

2.3

A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5
seconds. If the total car and driver mass is 1075 kg. Find the necessary force.
Solution:
Acceleration is the time rate of change of velocity.
ma = ∑ F ;

a = dV / dt = (60 × 1000) / (3600 × 5) = 3.33 m/s2

Fnet = ma = 1075 ×3.333 = 3583 N
2.4

A washing machine has 2 kg of clothes spinning at a rate that generates an
acceleration of 24 m/s2. What is the force needed to hold the clothes?
Solution:
F = ma = 2 kg × 24 m/s2 = 48 N

2.5

A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a
speed of 75 km/h. What are the force and total time required?
Solution:
a = dV / dt => ∆ t = dV/a = [ (75 − 20 ) / 4 ] × ( 1000 / 3600 )
∆ t = 3.82 sec ; F = ma = 1200 × 4 = 4800 N

2-3
2.6

A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What
force is needed and what is the final velocity?
Solution:
Constant acceleration can be integrated to get velocity.
a = dV / dt =>
V = 30 m/s ;

2.7

∫ dV = ∫ a dt

=> ∆ V = a ∆ t = 3 × 10 = 30 m/s

F = ma = 950 × 3 = 2850 N

A15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2
kN now accelerates this system. What is the acceleration?
Solution:
ma = ∑ F ⇒ a = ∑ F / m
m = msteel + mpropane = 15 + (1.75 × 44.094) = 92.165 kg
a = 2000 / 92.165 = 21.7 m/s2

2.8

A rope hangs over a pulley with the two equally long ends down. On one end you
attach a mass of 5 kg and on the other end you attach 10kg. Assuming standard
gravitation and no friction in the pulley what is the acceleration of the 10 kg mass
when released?
Solution:
Do the equation of motion for the mass m2 along the
downwards direction, in that case the mass m 1 moves
up (i.e. has -a for the acceleration)
m2 a = m2 g − m1 g − m1a
(m1 + m2 ) a = (m2 − m1 )g
This is net force in motion direction
a = (10 − 5) g / (10 + 5) = g / 3 =3.27 m/s2

2.9

g

2
1

A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration
of 2 m/s2 relative to the ground at a location where the local gravitational
acceleration is 9.5 m/s2. Find the required force.
Solution:
F = ma = Fup − mg
Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N

2-4
2.10

On the moon the gravitational acceleration is approximately one-sixth that on thesurface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface
on the moon. What is the expected reading? If this mass is weighed with a spring
scale that reads correctly for standard gravity on earth (see Problem 2.1), what is
Solution:
Moon gravitation is: g = gearth/6

m

m

m
2.11

Beam Balance Reading is 5 kg
This is mass comparison

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