According to Newton's First Law, a body will continue in its state of rest or uniform motion unless acted upon by a net external force. In this scenario, the astronaut is in interstellar space, far from significant gravitational influences from stars or the spaceship itself. Therefore, the net external force acting on the astronaut is zero.

Since the net external force is zero ($\mathbf{F}_{net} = 0$), Newton's First Law states that the acceleration of the astronaut must also be zero ($\mathbf{a} = 0$).

The acceleration of the astronaut the instant after he is outside the spaceship is zero. His velocity at that instant will be the same as the spaceship's velocity at that instant, and he will continue to move at that constant velocity.

Given mass of the bullet $m = 0.04 \text{ kg}$. Initial speed $v_0 = 90 \text{ m s}^{-1}$. Final speed $v = 0 \text{ m s}^{-1}$ (stopped). Distance traveled in the block $s = 60 \text{ cm} = 0.60 \text{ m}$.

We assume the resistive force (and thus the acceleration) is constant. We can use the kinematic equation $v^2 = v_0^2 + 2as$. Let the direction of motion be positive. The acceleration $a$ will be negative (retardation).

$0^2 = (90 \text{ m s}^{-1})^2 + 2a(0.60 \text{ m})$.

$0 = 8100 \text{ m}^2 \text{ s}^{-2} + 1.20 a \text{ m}$.

$1.20 a \text{ m} = -8100 \text{ m}^2 \text{ s}^{-2}$.

$a = \frac{-8100 \text{ m}^2 \text{ s}^{-2}}{1.20 \text{ m}} = -6750 \text{ m s}^{-2}$.

The magnitude of the acceleration is $6750 \text{ m s}^{-2}$, and it is in the direction opposite to motion.

According to Newton's Second Law, the net force $\mathbf{F} = m\mathbf{a}$. The resistive force $\mathbf{F}_r$ is the force exerted by the block on the bullet, and it causes the acceleration $a$. Assuming no other horizontal forces, $\mathbf{F}_r = m\mathbf{a}$.

$|\mathbf{F}_r| = m |\mathbf{a}| = (0.04 \text{ kg}) \times (6750 \text{ m s}^{-2})$.

$|\mathbf{F}_r| = 270 \text{ N}$.

The direction of the force is the same as the acceleration, opposite to the bullet's initial motion.

The average resistive force exerted by the block on the bullet is 270 N, in the direction opposite to the bullet's motion.

The position of the particle along the y-axis is given by $y(t) = ut + \frac{1}{2}gt^2$. Here, $u$ and $g$ are constants.

To find the force acting on the particle, we first need to find its acceleration. Acceleration is the second derivative of position with respect to time ($\mathbf{a} = \frac{d^2\mathbf{r}}{dt^2}$). In one dimension (along the y-axis), $a_y = \frac{d^2y}{dt^2}$.

First derivative (velocity): $v_y = \frac{dy}{dt} = \frac{d}{dt}(ut + \frac{1}{2}gt^2) = u + \frac{1}{2}g(2t) = u + gt$.

Second derivative (acceleration): $a_y = \frac{dv_y}{dt} = \frac{d}{dt}(u + gt) = 0 + g = g$.

So, the acceleration of the particle is constant and equal to $g$ in the direction of the y-axis.

According to Newton's Second Law, the net force $\mathbf{F} = m\mathbf{a}$. Since the motion is along the y-axis, the force is also along the y-axis, $\mathbf{F} = F_y \hat{\mathbf{j}}$.

$F_y = m a_y = mg$.

The force acting on the particle is $mg$ in the direction of the y-axis. This equation describes the motion of a particle of mass $m$ under a constant force $mg$ along the y-axis (which could be, for example, the force of gravity if the y-axis is chosen to be vertical and pointing downwards). $u$ is the initial velocity in the y-direction.

Given mass of the ball $m = 0.15 \text{ kg}$. Initial speed $v_i = 12 \text{ m s}^{-1}$. Final speed $v_f = 12 \text{ m s}^{-1}$ (speed does not change). The ball is hit back straight towards the bowler, so the direction of motion is reversed.

Let the direction of the bowler to the batsman be positive. Then the initial velocity is $\mathbf{v}_i = +12 \text{ m s}^{-1}$. Since the ball is hit back straight in the direction of the bowler, the final velocity is $\mathbf{v}_f = -12 \text{ m s}^{-1}$ (opposite direction).

Initial momentum $\mathbf{p}_i = m \mathbf{v}_i = (0.15 \text{ kg})(+12 \text{ m s}^{-1}) = +1.8 \text{ kg m s}^{-1}$.

Final momentum $\mathbf{p}_f = m \mathbf{v}_f = (0.15 \text{ kg})(-12 \text{ m s}^{-1}) = -1.8 \text{ kg m s}^{-1}$.

Impulse imparted to the ball is equal to the change in momentum $\mathbf{I} = \Delta \mathbf{p} = \mathbf{p}_f - \mathbf{p}_i$.

$\mathbf{I} = (-1.8 \text{ kg m s}^{-1}) - (+1.8 \text{ kg m s}^{-1}) = -3.6 \text{ kg m s}^{-1}$.

The magnitude of the impulse is 3.6 $\text{kg m s}^{-1}$ or 3.6 N s (units are equivalent). The negative sign indicates the direction of the impulse. Since we took the direction towards the batsman as positive, the negative impulse means the impulse is in the direction towards the bowler.

The impulse imparted to the ball is 3.6 N s, directed towards the bowler.

Let the mass of each ball be $m$ and the speed be $u$. Let the wall be along the y-axis, and the direction perpendicular to the wall (towards the wall) be the positive x-axis. The force exerted by the wall on the ball is the force that changes the ball's momentum. By Newton's Third Law, the force exerted by the ball on the wall is equal and opposite to the force by the wall on the ball.

The force on the ball by the wall is typically a normal force perpendicular to the wall surface during contact, unless there is also friction (not mentioned here, and usually assumed negligible in simple collision problems unless specified). In this case, since the wall is rigid and smooth (implied by no friction or damage), the force is normal to the wall.

(i) The direction of the force on the wall due to each ball: By Newton's Third Law, the force on the wall by the ball is equal and opposite to the force on the ball by the wall. The force by the wall on the ball is normal to the wall (along the x-axis). In case (a), the ball approaches along the +x direction and rebounds along the -x direction. The change in momentum is primarily in the x-direction, so the force on the ball is in the -x direction. The force on the wall is in the +x direction (away from the wall). In case (b), the x-component of velocity changes direction. The y-component of velocity does not change (speed remains the same, angle of reflection equals angle of incidence with respect to normal). The force by the wall on the ball changes the x-component of momentum, so this force is purely in the -x direction. By the third law, the force on the wall is in the +x direction.

In both cases, the force on the wall due to the ball is normal to the wall, directed away from the wall.

(ii) The ratio of the magnitudes of impulses imparted to the balls by the wall? Impulse on the ball by the wall is the change in momentum of the ball. $\mathbf{I} = \Delta \mathbf{p} = \mathbf{p}_f - \mathbf{p}_i = m\mathbf{v}_f - m\mathbf{v}_i$.

Case (a): The ball strikes normally. Let the initial velocity be $\mathbf{v}_i = u \hat{\mathbf{i}}$. The reflected velocity is $\mathbf{v}_f = -u \hat{\mathbf{i}}$.

Impulse $\mathbf{I}_a = m(-u \hat{\mathbf{i}}) - m(u \hat{\mathbf{i}}) = -2mu \hat{\mathbf{i}}$. Magnitude $|\mathbf{I}_a| = 2mu$.

Case (b): The ball strikes at 30° with the normal. Let the initial velocity be $\mathbf{v}_i = u \cos 30^\circ \hat{\mathbf{i}} - u \sin 30^\circ \hat{\mathbf{j}}$. The reflected velocity is $\mathbf{v}_f = -u \cos 30^\circ \hat{\mathbf{i}} - u \sin 30^\circ \hat{\mathbf{j}}$. The vertical component ($y$) does not change sign upon reflection from the wall, while the horizontal component ($x$) does.

Impulse $\mathbf{I}_b = m\mathbf{v}_f - m\mathbf{v}_i = m(-u \cos 30^\circ \hat{\mathbf{i}} - u \sin 30^\circ \hat{\mathbf{j}}) - m(u \cos 30^\circ \hat{\mathbf{i}} - u \sin 30^\circ \hat{\mathbf{j}})$

$\mathbf{I}_b = m(-u \cos 30^\circ \hat{\mathbf{i}} - u \sin 30^\circ \hat{\mathbf{j}} - u \cos 30^\circ \hat{\mathbf{i}} + u \sin 30^\circ \hat{\mathbf{j}})$

$\mathbf{I}_b = m(-2u \cos 30^\circ \hat{\mathbf{i}})$. The impulse is purely in the x-direction, normal to the wall, which is consistent with our earlier conclusion about the force direction.

Magnitude $|\mathbf{I}_b| = |-2mu \cos 30^\circ| = 2mu \cos 30^\circ = 2mu (\sqrt{3}/2) = \sqrt{3}mu$.

The ratio of the magnitudes of impulses $|\mathbf{I}_a|/|\mathbf{I}_b| = (2mu)/(\sqrt{3}mu) = 2/\sqrt{3}$.

The ratio of the magnitudes of impulses imparted to the balls by the wall in case (a) to case (b) is $2/\sqrt{3}$.

The system is in equilibrium, so the net force on every part of the system is zero. We can analyze the forces acting on the mass W and the point P separately using free-body diagrams.

Consider the mass W (6 kg). It is in equilibrium, so the net force on it is zero. The forces on W are:

• Its weight, $W = mg = (6 \text{ kg})(10 \text{ m s}^{-2}) = 60 \text{ N}$, vertically downwards.
• The tension in the lower part of the rope, $T_2$, vertically upwards.

For equilibrium of W: $T_2 - W = 0 \implies T_2 = W = 60 \text{ N}$.

Now consider the point P, the midpoint of the rope. Neglecting the mass of the rope means P is essentially a point where the forces are concurrent. The forces acting on point P are:

• The tension in the upper part of the rope, $T_1$, acting upwards at an angle $\theta$ with the vertical.
• The tension in the lower part of the rope, $T_2$, acting downwards (equal to the tension pulling up the mass W). $T_2 = 60 \text{ N}$.
• The horizontal force $F = 50 \text{ N}$, acting horizontally.

For equilibrium of point P, the net force must be zero. Resolve $T_1$ into horizontal and vertical components. Let the positive x-direction be horizontal (in the direction of F) and the positive y-direction be vertical upwards.

• Horizontal forces: $F - T_1 \sin \theta = 0 \implies T_1 \sin \theta = 50$ N.
• Vertical forces: $T_1 \cos \theta - T_2 = 0 \implies T_1 \cos \theta = T_2 = 60$ N.

We have two equations: $T_1 \sin \theta = 50$ and $T_1 \cos \theta = 60$.

Divide the first equation by the second: $\frac{T_1 \sin \theta}{T_1 \cos \theta} = \frac{50}{60}$.

$\tan \theta = \frac{5}{6}$.

$\theta = \arctan(5/6) \approx \arctan(0.8333) \approx 39.8^\circ$.

The angle the rope makes with the vertical in equilibrium is approximately 39.8°.

We can also find $T_1$: Square both equations and add them: $(T_1 \sin \theta)^2 + (T_1 \cos \theta)^2 = 50^2 + 60^2$.

$T_1^2 (\sin^2 \theta + \cos^2 \theta) = 2500 + 3600$.

$T_1^2 (1) = 6100$.

$T_1 = \sqrt{6100} \approx 78.1 \text{ N}$.

Let the mass of the box be $m$. The box remains stationary relative to the train, which means the box is accelerating with the same acceleration as the train, say $a$. According to Newton's Second Law ($\mathbf{F}_{net} = m\mathbf{a}$), there must be a net force on the box causing this acceleration.

The forces acting on the box are its weight ($mg$) downwards, the normal force ($N$) from the floor upwards, and the static frictional force ($f_s$) exerted by the floor in the horizontal direction (the direction of the train's acceleration).

In the vertical direction, there is no vertical acceleration, so $N - mg = 0 \implies N = mg$.

In the horizontal direction, the net force is the static friction: $f_s = ma$.

The static friction force can only provide acceleration up to its maximum limit: $f_s \le \mu_s N$.

Substitute the expression for $f_s$ and $N$ into the inequality:

$ma \le \mu_s (mg)$.

Cancel mass $m$ from both sides:

$a \le \mu_s g$.

The maximum acceleration the box can have (and remain stationary with respect to the train) is when the static friction is at its maximum value. Thus, the maximum acceleration of the train is $a_{max} = \mu_s g$.

Given $\mu_s = 0.15$ and $g = 10 \text{ m s}^{-2}$ (as instructed in the problem set header):

$a_{max} = 0.15 \times 10 \text{ m s}^{-2} = 1.5 \text{ m s}^{-2}$.

The maximum acceleration of the train in which the box will remain stationary is 1.5 m s⁻².

Let the mass of the block be $m = 4 \text{ kg}$. The block is on an inclined plane. The forces acting on the block are its weight ($mg$) vertically downwards, the normal force ($N$) perpendicular to the plane upwards, and the static frictional force ($f_s$) parallel to the plane, opposing the impending downward motion.

Resolve the weight $mg$ into components parallel and perpendicular to the inclined plane. The angle of inclination is $\theta$.

• Component of weight perpendicular to the plane: $mg \cos \theta$, acting downwards into the plane.
• Component of weight parallel to the plane: $mg \sin \theta$, acting downwards along the plane.

The block is just beginning to slide at $\theta = 15^\circ$, meaning it is in equilibrium, and the static friction is at its maximum value, $(f_s)_{max}$.

For equilibrium perpendicular to the plane: $N - mg \cos \theta = 0 \implies N = mg \cos \theta$.

For equilibrium parallel to the plane: $f_s - mg \sin \theta = 0 \implies f_s = mg \sin \theta$.

At the angle where sliding just begins ($\theta = 15^\circ$), the static friction reaches its maximum value, $(f_s)_{max} = \mu_s N$.

So, $mg \sin 15^\circ = (f_s)_{max} = \mu_s N = \mu_s (mg \cos 15^\circ)$.

Cancel $mg$ from both sides:

$\sin 15^\circ = \mu_s \cos 15^\circ$.

$\mu_s = \frac{\sin 15^\circ}{\cos 15^\circ} = \tan 15^\circ$.

The coefficient of static friction $\mu_s$ is equal to the tangent of the angle of inclination at which the block just starts to slide. This angle is called the angle of repose.

$\mu_s = \tan 15^\circ$.

From trigonometric tables or calculator, $\tan 15^\circ \approx 0.2679$.

The coefficient of static friction between the block and the surface is approximately 0.27.

The system consists of a 3 kg block hanging vertically, connected by a string over a pulley to a 20 kg trolley on a horizontal surface. Since the string is light and inextensible and the pulley is smooth, both the block and the trolley will move with the same magnitude of acceleration, $a$. Let the tension in the string be $T$. The 3 kg block will accelerate downwards, and the 20 kg trolley will accelerate horizontally.

Consider the free-body diagram of the 3 kg block. Forces on the block are its weight ($m_1 g$) downwards and the tension ($T$) upwards.

Applying Newton's Second Law ($F_{net} = ma$) in the vertical direction (downwards positive):

$m_1 g - T = m_1 a$

$(3 \text{ kg})(10 \text{ m s}^{-2}) - T = (3 \text{ kg}) a$

$30 - T = 3a$ (Equation 1)

Consider the free-body diagram of the 20 kg trolley. Forces on the trolley are its weight ($m_2 g$) downwards, the normal force ($N$) from the surface upwards, the tension ($T$) in the string pulling horizontally, and the kinetic frictional force ($f_k$) opposing the horizontal motion.

In the vertical direction, there is no vertical acceleration, so $N - m_2 g = 0 \implies N = m_2 g = (20 \text{ kg})(10 \text{ m s}^{-2}) = 200 \text{ N}$.

The kinetic frictional force is $f_k = \mu_k N$. Given $\mu_k = 0.04$.

$f_k = 0.04 \times 200 \text{ N} = 8 \text{ N}$.

Applying Newton's Second Law ($F_{net} = ma$) in the horizontal direction (in the direction of acceleration):

$T - f_k = m_2 a$

$T - 8 = 20 a$ (Equation 2)

Now we have a system of two linear equations with two unknowns ($a$ and $T$):

1) $30 - T = 3a$

2) $T - 8 = 20a$

Add Equation 1 and Equation 2 to eliminate $T$:

$(30 - T) + (T - 8) = 3a + 20a$

$22 = 23a$

$a = \frac{22}{23} \text{ m s}^{-2}$.

$a \approx 0.9565 \text{ m s}^{-2}$.

The acceleration of the system is approximately 0.96 m s⁻².

Substitute the value of $a$ back into Equation 2 to find $T$:

$T - 8 = 20a = 20 \times \frac{22}{23} = \frac{440}{23} \text{ N}$.

$T = 8 + \frac{440}{23} = \frac{8 \times 23 + 440}{23} = \frac{184 + 440}{23} = \frac{624}{23} \text{ N}$.

$T \approx 27.13 \text{ N}$.

The tension in the string is approximately 27.1 N.

The cyclist is taking a circular turn on a level road. The centripetal force needed is provided by the static friction between the tyres and the road. The maximum static friction available limits the maximum speed for the turn.

Given speed $v = 18 \text{ km/h}$. Convert to m/s: $v = 18 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m s}^{-1}$.

Radius of the turn $R = 3 \text{ m}$.

Coefficient of static friction $\mu_s = 0.1$.

Acceleration due to gravity $g = 9.8 \text{ m s}^{-2}$ (using the standard value unless instructed otherwise for this problem). Let's use $g=10$ as per chapter instruction.

The condition for not slipping is that the required centripetal force ($mv^2/R$) must be less than or equal to the maximum static friction ($\mu_s N$). On a level road, $N = mg$.

$\frac{mv^2}{R} \le \mu_s mg$.

Cancel mass $m$ from both sides:

$\frac{v^2}{R} \le \mu_s g$.

Rearrange to find the maximum speed allowed by friction: $v_{max}^2 = \mu_s Rg$, so $v_{max} = \sqrt{\mu_s Rg}$.

$v_{max} = \sqrt{0.1 \times 3 \text{ m} \times 10 \text{ m s}^{-2}} = \sqrt{3 \text{ m}^2 \text{ s}^{-2}} = \sqrt{3} \text{ m s}^{-1}$.

$v_{max} \approx 1.732 \text{ m s}^{-1}$.

The cyclist's speed is $v = 5 \text{ m s}^{-1}$, and the maximum speed they can take this turn without slipping is only about 1.732 m s⁻¹. Since the cyclist's speed is much greater than the maximum permissible speed ($5 > 1.732$), the required centripetal force ($mv^2/R$) is much greater than the maximum static friction available ($\mu_s mg$).

Therefore, the cyclist will slip while taking the turn.

Given radius $R = 300 \text{ m}$. Banking angle $\theta = 15^\circ$. Coefficient of friction $\mu_s = 0.2$. Acceleration due to gravity $g = 10 \text{ m s}^{-2}$ (as per chapter instruction).

(a) The optimum speed ($v_0$) is the speed at which the centripetal force is provided solely by the horizontal component of the normal force, and friction is zero. This speed is given by the formula $v_0 = \sqrt{Rg \tan \theta}$.

$v_0 = \sqrt{(300 \text{ m})(10 \text{ m s}^{-2})\tan 15^\circ}$.

$\tan 15^\circ \approx 0.2679$.

$v_0 = \sqrt{3000 \times 0.2679} = \sqrt{803.7} \text{ m/s}$.

$v_0 \approx 28.35 \text{ m/s}$.

Converting to km/h: $v_0 = 28.35 \text{ m/s} \times \frac{3600 \text{ s}}{1000 \text{ m}} = 28.35 \times 3.6 \text{ km/h} \approx 102.06 \text{ km/h}$.

The optimum speed is approximately 28.4 m/s (or 102 km/h).

(b) The maximum permissible speed ($v_{max}$) is the highest speed the car can take the turn without slipping, considering the maximum static friction acting downwards along the inclined road (aiding the banking). This speed is given by the formula $v_{max} = \sqrt{Rg \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}}$.

$v_{max} = \sqrt{(300 \text{ m})(10 \text{ m s}^{-2}) \frac{\tan 15^\circ + 0.2}{1 - (0.2)(\tan 15^\circ)}}$.

$v_{max} = \sqrt{3000 \frac{0.2679 + 0.2}{1 - (0.2)(0.2679)}} = \sqrt{3000 \frac{0.4679}{1 - 0.05358}}$.

$v_{max} = \sqrt{3000 \frac{0.4679}{0.94642}} = \sqrt{3000 \times 0.49443}$.

$v_{max} = \sqrt{1483.3} \text{ m/s}$.

$v_{max} \approx 38.51 \text{ m/s}$.

Converting to km/h: $v_{max} = 38.51 \text{ m/s} \times 3.6 \approx 138.6 \text{ km/h}$.

The maximum permissible speed is approximately 38.5 m/s (or 139 km/h).

Given: mass of block $m_b = 2 \text{ kg}$. Mass of cylinder $m_c = 25 \text{ kg}$. Acceleration due to gravity $g = 10 \text{ m s}^{-2}$.

(a) Action of the block on the floor before the floor yields:

Before the cylinder is placed, the block of mass 2 kg is at rest on the floor. The forces on the block are its weight ($W_b$) downwards and the normal force ($N_{floor, b}$) from the floor upwards. Since the block is at rest ($\mathbf{a} = 0$), the net force on it is zero by Newton's First Law.

$N_{floor, b} - W_b = 0 \implies N_{floor, b} = W_b = m_b g = (2 \text{ kg})(10 \text{ m s}^{-2}) = 20 \text{ N}$.

By Newton's Third Law, the force exerted by the block on the floor (action) is equal in magnitude and opposite in direction to the force exerted by the floor on the block (normal force, reaction). The normal force is 20 N upwards.

The action of the block on the floor is 20 N vertically downwards.

(b) Action of the block on the floor after the floor yields:

Now the iron cylinder is on top of the block, and the system (block + cylinder) accelerates downwards with $a = 0.1 \text{ m s}^{-2}$.

Consider the system as the combined mass of the block and cylinder ($M_{sys} = m_b + m_c = 2 \text{ kg} + 25 \text{ kg} = 27 \text{ kg}$). The forces on this system are its total weight ($W_{sys}$) downwards and the normal force ($R'_{floor, sys}$) from the floor upwards.

$W_{sys} = M_{sys} g = (27 \text{ kg})(10 \text{ m s}^{-2}) = 270 \text{ N}$, vertically downwards.

The system is accelerating downwards with $a = 0.1 \text{ m s}^{-2}$. Let's apply Newton's Second Law to the system in the downward direction (downwards positive):

$W_{sys} - R'_{floor, sys} = M_{sys} a$

$270 \text{ N} - R'_{floor, sys} = (27 \text{ kg})(0.1 \text{ m s}^{-2})$

$270 - R'_{floor, sys} = 2.7 \text{ N}$.

$R'_{floor, sys} = 270 - 2.7 = 267.3 \text{ N}$.

This is the normal force exerted by the floor upwards on the combined system. By Newton's Third Law, the action of the combined system (block + cylinder) on the floor is equal and opposite to this reaction force.

The action of the combined system on the floor is 267.3 N vertically downwards.

The question asks for the action of the *block* on the floor. This is subtle. When the cylinder is on the block, the block exerts a force on the floor. This force is the normal force the floor provides to the block. The block is supporting the cylinder *and* its own weight while accelerating. The forces on the block are its weight ($W_b$), the force from the cylinder ($F_{c \to b}$) downwards, and the normal force from the floor ($N'_{floor, b}$) upwards. The block is accelerating downwards with $a=0.1 \text{ m s}^{-2}$.

$W_b + F_{c \to b} - N'_{floor, b} = m_b a$ (downwards positive)

$20 \text{ N} + F_{c \to b} - N'_{floor, b} = (2 \text{ kg})(0.1 \text{ m s}^{-2}) = 0.2 \text{ N}$.

We need $F_{c \to b}$. Consider the cylinder's free-body diagram. Forces on the cylinder are its weight ($W_c$) downwards and the normal force from the block upwards ($F_{b \to c}$). The cylinder accelerates downwards with $a=0.1 \text{ m s}^{-2}$.

$W_c - F_{b \to c} = m_c a$ (downwards positive)

$(25 \text{ kg})(10 \text{ m s}^{-2}) - F_{b \to c} = (25 \text{ kg})(0.1 \text{ m s}^{-2})$

$250 \text{ N} - F_{b \to c} = 2.5 \text{ N}$.

$F_{b \to c} = 250 - 2.5 = 247.5 \text{ N}$.

By Newton's Third Law, the force of the cylinder on the block ($F_{c \to b}$) is equal and opposite to the force of the block on the cylinder ($F_{b \to c}$). So, $F_{c \to b} = 247.5 \text{ N}$ downwards on the block.

Now go back to the block's equation:

$20 \text{ N} + 247.5 \text{ N} - N'_{floor, b} = 0.2 \text{ N}$.

$267.5 \text{ N} - N'_{floor, b} = 0.2 \text{ N}$.

$N'_{floor, b} = 267.5 - 0.2 = 267.3 \text{ N}$.

This is the normal force exerted by the floor upwards on the block. By Newton's Third Law, the action of the block on the floor is equal and opposite to this reaction force.

The action of the block on the floor is 267.3 N vertically downwards.

Notice that the action of the block on the floor in case (b) is equal to the total weight of the system minus the net force causing the downward acceleration ($W_{sys} - M_{sys} a = 270 - 27 \times 0.1 = 270 - 2.7 = 267.3 \text{ N}$). This makes sense as the floor is supporting less than the total weight because the system is accelerating downwards.

Action-reaction pairs:

• Weight of block on Earth (Action) / Force of Earth on block (Reaction) - 20 N pair.
• Weight of cylinder on Earth (Action) / Force of Earth on cylinder (Reaction) - 250 N pair.
• Force of block on floor (Action) / Force of floor on block (Reaction) - 20 N pair in (a), 267.3 N pair in (b).
• Force of cylinder on block (Action) / Force of block on cylinder (Reaction) - 247.5 N pair in (b).

An action-reaction pair involves forces acting on two different bodies. For example, the weight of the block (force by Earth on block) and the normal force on the block by the floor are NOT an action-reaction pair; they are two forces acting on the same body (the block).