a)

First Newton's law

If no force or forces act on the body, the body remains at rest or moves in a uniform linear motion relative to the inertial reference frame.

Second Newton's law

If the body has a constant unbalanced forceunbalanced forceunbalanced force (or net force), the body moves in a uniformly accelerated motion with acceleration directly proportional to the applied force and inversely proportional to the mass of this body.

Third Newton's law

When the body A acts on the body B with a certain force ${\overrightarrow{F}}_{AB}$, the body B acts on the body A with the force ${\overrightarrow{F}}_{BA}$ with the same magnitude but opposite direction. These forces cannot be balanced because they are applied to two different bodies.

The third Newton's law is often called action‑reaction law.

b) The measure of body's inertia is its mass.

c) We distinguish two basic types of frictionfrictionfriction: static friction and dynamic friction.

d) The friction force depends on how large is the force which presses the two surfaces together and on the type of the two surfaces.

e) The friction force can be changed by varying the force which presses the surfaces together or by changing the coefficient of friction by changing the roughness of the surfaces (e.g. by smoothing) or by using a lubricant to reduce friction, grease or oil.

Analysis

Given:

$\mathrm{m} = 2 \mathrm{kg}$
$\mathrm{F} = 20 \mathrm{N}$
${\mathrm{m}}_1 = 0,5 \mathrm{kg}$
${\mathrm{F}}_1 = 5 \mathrm{N}$

Unknown:

${\mathrm{a}}_1 = ?$

${\mathrm{a}}_2 = ?$

Solution:

$a=\frac{F}{m}$

$a=\frac{20N}{2kg}$

$a=10\frac{N}{kg}$

$a_1=\frac{F_1}{m_1}$

$a_1=\frac{5N}{0,5kg}$

$a_1=10\frac{kg\cdot \frac{m}{s^2}}{kg}=10\frac{m}{s^2}$

Answer:

Acceleration with which the body falls to Earth is $10\frac{m}{s^2}$, it is gravitational acceleration.

Analysis

Given:

${\mathrm{F}}_1 = 10 \mathrm{N}$
${\mathrm{F}}_2 = 20 \mathrm{N}$
${\mathrm{F}}_3 = 30 \mathrm{N}$
$\mathrm{m} = 10 \mathrm{kg}$

Unknown:

$\mathrm{a} = ?$

Solution:

$a=\frac{F_w}{m}$

$F_w=F_2+F_3-F_1$

$F_w=20N+30N-10N$

$F_w=40N$

$a=4\frac{kg\cdot \frac{m}{s^2}}{kg}=4\frac{m}{s^2}$

Answer:

The block acceleration is $4\frac{m}{s^2}$.

Analysis

Given:

$\mathrm{m} = 200 \mathrm{kg}$
$\mathrm{a} = 2,5 \frac{\mathrm{m}}{{\mathrm{s}}^2}$ 
$\mathrm{g} = 10 \frac{\mathrm{m}}{{\mathrm{s}}^2}$

Unknown:

${\mathrm{F}}_{\mathrm{T}} = ?$
${\mathrm{F}}_{\mathrm{N}} = ?$

Solution:

$a=\frac{F}{m}$

$F=a\cdot m$

$F_T=F=2,5\frac{m}{s^2}\cdot 200kg$

$F=500N$

$F_g=F_N=m\cdot g$

$F_N=200kg\cdot 10\frac{m}{s^2}$

$F_N=2000N$

$\frac{F_T}{F_N}=\frac{500N}{2000N}=\frac{1}{4}$

$F_T=\frac{1}{4}{F}_N$

Answer:

The magnitude of the frictionfrictionfriction force is 500 N. The friction force is $\frac{1}{4}$ of the normal force.

a) True. 
b) False. 
c) True. 
d) True.

a) False. 
b) True. 
c) False. 
d) True.

a) False. 
b) False. 
c) True. 
d) True. 
c) True.

a) True. 
b) True. 
c) False. 
d) False. 
e) False. 
f) False.

a) False. 
b) True. 
c) False. 
d) False. 
e) False.

The net force F setting the blocks in motion is equal to the difference in weight of the hanging blocks:

$\mathrm{F} = 4 \mathrm{kg} · 10 \frac{\mathrm{m}}{{\mathrm{s}}^2} - 2 \mathrm{kg} · 10 \frac{\mathrm{m}}{{\mathrm{s}}^2}= 20 \mathrm{kg} \frac{\mathrm{m}}{{\mathrm{s}}^2}$

The mass m on which the force F acts is the sum of three masses:

$\mathrm{m} = 2 \mathrm{kg} + 2 \mathrm{kg} + 4 \mathrm{kg} = 8 \mathrm{kg}$

According to the second Newton’s law, the acceleration of the three masses is:

a = $\frac{{\displaystyle F}}{m}$

a = $\frac{20kg\frac{m}{s^2}}{8kg}$

a = 2,5 $\frac{m}{s^2}$

Answer:

a = 2,5 $\frac{m}{s^2}$

In this phenomenon, besides the locomotive and the wagon, there is also the Earth. In this system, the locomotive, by means of friction (kinetic) of wheels on rails, pushes them together with the Earth backwards, while the Earth pushes the locomotive forward. The weight of the locomotive, which is bigger than the weight of the wagon, is significant here, hence the friction force of the locomotive's wheels on the rails is greater than the friction force of the wagon wheels on the rails. Therefore, when starting off, the resultant of forces acting on the train is directed towards the motion of the locomotive. With uniform motion, all forces acting on the train are balanced, but the inertia of the train results in maintaining the acquired direction of movement.