Newton's Laws

Both wagons will travel the same distance.

$m = 2(\frac{F}{a} - M) \approx 800$ kg.

The conditions for uniform motion of the aerostat during descent and ascent are expressed respectively: $Mg = F + F_{\text{res}}$; $(M-m)g + F_{\text{res}} = F$.

a) $F_N = M(g+a) \approx 3.1 \times 10^4$ H;

b) $F_N = M(g-a) \approx 2.8 \times 10^4$ H;

c) $F_N = Mg \approx 2.95 \times 10^4$ H.

When the elevator moves with acceleration directed upwards (this will be at the beginning of the ascent and at the end of the descent), the scale reading is greater than $mg$ ($m$ is the mass of the person). If the acceleration is directed downwards, which will be at the beginning of the descent and at the end of the ascent, the scale reading is less than $mg$. If the elevator moves uniformly, the scale reading is equal to $Mg$.

The equation of motion for the person is $ma = mg - N$ (equation written in projections onto the direction of $g$). Here $a$ is the acceleration of the elevator (algebraic quantity) and $N$ is the reaction force of the scales. In accordance with Newton's third law, the force $N'$ acting on the scales from the person is equal in magnitude to $N$: $N' = N = m(g-a)$.

When squatting, it will decrease; when straightening up, it will increase.

When squatting, the acceleration of the person's center of mass is directed downwards, and when straightening up - upwards.

During the free movement of the box (movement in the field of gravity in the absence of other forces), the pressure force of the balls on the bottom and walls of the box, as well as on each other, is zero.

The body will oscillate on the spring, the amplitude of which is equal to the initial extension of the spring.

The problem is conveniently solved in a non-inertial coordinate system associated with the moving elevator (see 'Elementary Physics Textbook' edited by Acad. G. S. Landsberg, vol. I, ch. 6). In this reference frame, three forces act on the mass: the force of gravity $mg$, the inertial force $-ma$ balancing it, and also the spring tension force, equal at the initial moment to the force of gravity $mg$ acting on the mass.

$\Delta t = \sqrt{2m(g-a)/(ka)}$.

Until the moment the body detaches from the support, it moves with acceleration $a$ under the action of three forces: the force of gravity $F_1 = mg$, the spring tension force $F_2 = kx$, and the support reaction force $N$. Therefore $mg - kx - N = ma$. At the moment the body detaches from the support, the support reaction force becomes zero. Considering this and substituting the expression for $x = a(\Delta t)^2/2$ into the equation of motion, we find $\Delta t = \sqrt{2m(g-a)/(ka)}$.

$a \approx 1.96$ m/s$^2$; $T \approx 0.6$ H; $F \approx 1.2$ H.

The forces acting on the weights are shown in the following figure. Let's write the equations of motion for the weights: $m_1 a_1 = m_1 g - T_1,\tag{1}$ $m_2 a_2 = T_2 - m_2 g. \tag{2}$ The accelerations $a_1$ and $a_2$ of the weights are numerically equal, since the string is inextensible and the displacements of the weights are always the same (the relationships between the accelerations of the bodies of the system are determined by their kinematic connection, i.e., the connection between their displacements). In the general case, the forces $T_1$ and $T_2$ applied to the ends of the string, according to Newton's third law, are numerically equal to the forces $T'_1$ and $T'_2$, are not equal to each other, i.e., the difference between these forces imparts acceleration to the string and angular acceleration to the pulley. If the mass of the string and pulley can be neglected, then $T_1$ and $T_2$ and the corresponding forces $T'_1$ and $T'_2$ applied to the weights can be considered equal to each other. Projecting the vectors included in equations (1) and (2) onto the directions $x_1$ and $x_2$, coinciding with the accelerations of each of the weights (conventionally assume $m_1 > m_2$), we obtain the following system of equations: $m_1 a = m_1 g - T$; $m_2 a = T - m_2 g$. Solving this system for the unknowns $a$ and $T$, we get $a = \frac{m_1 - m_2}{m_1 + m_2} g$; $T = \frac{2m_1 m_2}{m_1 + m_2} g$. (If $m_1 < m_2$, we get $a < 0$, i.e., the weights move in the direction opposite to the assumed one.) The dynamometer reading is obviously equal to the sum of the string tension forces: $F = 2T = \frac{4m_1 m_2}{m_1 + m_2} g.$

$t = 0.21$ s.

See the previous problem. Each weight will travel a path equal to 5 cm.

$F = 4Mg (\dfrac{m+M}{m+2M})$; $f = 2 \dfrac{Mm}{m+2M} g$.

In accordance with Newton's second law, let's write the equations of motion for each of the bodies (see figure): for the first body $Ma = T - Mg;$ for the second body $Ma = Mg + f - T;$ for the load $ma = mg - f',$ where $T$ is the string tension force; $f$ is the pressure force of the load on the body, $a$ is the acceleration of the bodies; $f'$ is the reaction force of the support. According to Newton's third law, $f'$ and $f$, $T$ and $T'$ are numerically equal. The tensions on the right and left are equal, since friction and the mass of the pulley are neglected. Solving the obtained system of equations, we find $a = \frac{mg}{m+2M};$ $f = 2 \frac{Mm}{m+2M} g,$ $T = M(\frac{mg}{m+2M} + g) = 2M (\frac{m+M}{m+2M}) g.$ The force acting on the axis of the pulley, $F = 2T = 4M (\frac{m+M}{m+2M}) g.$

1. $F = \dfrac{4m_1 m_2}{m_1 + m_2} g \approx 64$ N.

2. $F = \frac{9m_1 m_2}{4m_1 + m_2} g \approx 42.3$ N.

1. The force $F'$, acting from the axis on the rod and according to Newton's third law numerically equal to the pressure force $F$ on the axis (see figure below, a), can be replaced by two forces $F_1$ and $F_2$, applied to each of the weights and equal to $F/2$. Considering that the rod is massless, we can, using Newton's second law, write: $m_1 a = F/2 - m_1 g$, $m_2 a = m_2 g - F/2$, whence $F = \frac{4m_1 m_2}{m_1 + m_2} g$.

2. The force $F'$ must be replaced (see figure below, b) by forces $F_1 = (2/3)F'$ and $F_2 = (1/3)F'$, applied to weights $m_1$ and $m_2$ respectively. The accelerations of the weights are related by $a_1/l_1 = a_2/l_2$, i.e., $a_1 = a_2/2$.

$a = \dfrac{m_1 \sin \alpha - m_2 \sin \beta}{m_1 + m_2} g$; $T = \dfrac{m_1 m_2 g (\sin \alpha + \sin \beta)}{m_1 + m_2}$.

Let's write the equations of Newton's second law for each of the weights, projecting the forces acting on the weights (see figure) onto the direction of possible motion of the corresponding weights, i.e., along the faces of the prism (assume that weight $m_1$ descends): $m_1 a = m_1 g \sin \alpha - T,$ $m_2 a = T - m_2 g \sin \beta.$ Solving this system, we find: $a = \frac{m_1 \sin \alpha - m_2 \sin \beta}{m_1 + m_2} g;$ $T = \frac{m_1 m_2 (\sin \alpha + \sin \beta)}{m_1 + m_2} g.$ If, upon substitution of the numerical values of masses and angles, it turns out that $a < 0$, it means that the entire system moves with acceleration $a$ in the direction opposite to the chosen one. The latter follows from the fact that in the absence of friction, changing the chosen direction of motion of the weights will only change the signs of the right sides of the equations derived above.

$a = 0.12$ m/s$^2$; $T = 0.22$ H; $T = \dfrac{F_1 M + F_2 m}{M+m} \approx \dfrac{F_1 + F_2 \frac{m}{M}}{1 + \frac{m}{M}} \approx F_1$, when $m \ll M$.

$a_1 = 2 \dfrac{2m_1 - m_2}{4m_1 + m_2} g$; $a_2 = \dfrac{2m_1 - m_2}{4m_1 + m_2} g$; $T = \dfrac{3m_1 m_2}{4m_1 + m_2} g$.

From the kinematic connection, it follows that the acceleration of weight $m_1$ is twice the acceleration of weight $m_2$ (when weight $m_1$ moves by distance $h$, weight $m_2$ moves by distance $h/2$). A force equal to $2T$ is applied to weight $m_2$ from the strings.

$\alpha = \arctan(a/g)$.

Let's conditionally isolate a small part of the liquid inside. Two forces act on this element of liquid with mass $m$: the force of gravity $mg$ and the buoyant (Archimedean) force $N$ from the side of the liquid. The resultant $F$ of these forces must be directed towards the acceleration of the vessel and equal to $ma$ (see figure). Therefore, the force $N$ must form an angle $\alpha = \arctan(a/g)$ with the vertical. The direction of the buoyant force $N$ acting on any isolated volume inside the liquid is perpendicular to the planes of equal pressure ($p$=const) in the liquid, is determined by them and, in particular, is perpendicular to its surface ($p=0$). The requirement of perpendicularity uniquely relates to the condition of rest of the liquid relative to the vessel. Any violation of perpendicularity is 'automatically' eliminated due to the fluidity of the liquid.

The surface of the liquid is parallel to the inclined plane.

1st method. See solution to problem 16.

2nd method. Let's solve the problem in the reference frame associated with the moving vessel (in a non-inertial system). In this case, forces $mg$, $N$, and the inertial force $|-m\vec{a}| = ma$ act on any element of the liquid, directed opposite to the motion of the vessel (see figure). The resultafignt of these forces must be zero (since the liquid is stationary relative to the vessel). This is possible when the force $N$ is perpendicular to the inclined plane. The surface of the liquid is parallel to it.

Will not change.

When the elevator moves with acceleration $\vec{a}$, Newton's second law for a floating body is written as: $m\vec{a} = m\vec{g} + \vec{F},\tag{1}$ where $\vec{F}$ is the buoyant force; $m$ is the mass of the body.

If the volume displaced by the body were occupied by liquid, a force, also equal to $\vec{F}$, would act on it from the rest of the liquid. According to Newton's second law
$m_l \vec{a} = m_l \vec{g} + \vec{F}; \tag{2}$
here $m_l = \rho_l V$, where $V$ is the volume of the submerged part of the body. Substituting the expression for $\vec{F}$ found from equality (1) into equation (2), we get
$m(\vec{a} - \vec{g}) = \rho_l V (\vec{a} - \vec{g}),$
hence it follows that $V = m/\rho_l$ does not depend on the acceleration of the elevator, i.e., the immersion depth of the body does not change. Note. The solution to the problem becomes very simple in the reference frame associated with the elevator (see problem 17).

$a = g \sin \alpha (1 + M/m)$; $k > (M/m) \tan \alpha$.

The dog must run with such acceleration, directed downwards, so that the force $-F_1$, with which the dog pushes the board back, balances the resultant $F_1$ of the forces acting on the board: gravity $Mg$ and reaction force $N_1$, i.e., equals $Mg \sin \alpha$ (see figure). On the dog, therefore, in the direction of its acceleration, act the force $mg \sin \alpha$ and the force $F_1$ from the board, equal, according to Newton's third law, to $Mg \sin \alpha$ (the direction of the dog's movement does not matter). The equation of motion for the dog will be $ma = mg \sin \alpha + Mg \sin \alpha$, whence $a = g \sin \alpha (1 + M/m)$. The force $-F_1$, with which the dog pushes the board, by its nature, is a friction force, which cannot exceed the value $kN$. The problem has a solution (the dog stops the board) if $kN \ge F_1$, i.e., $kmg \cos \alpha \ge Mg \sin \alpha$. Hence $k \ge (M/m) \tan \alpha$.

1st method. See solution to problem 16.

2nd method. Let's solve the problem in the reference frame associated with the moving vessel (in a non-inertial system). In this case, forces $mg$, $N$, and the inertial force $|-m\vec{a}| = ma$ act on any element of the liquid, directed opposite to the motion of the vessel (see figure). The resultafignt of these forces must be zero (since the liquid is stationary relative to the vessel). This is possible when the force $N$ is perpendicular to the inclined plane. The surface of the liquid is parallel to it.

$Mg$.

The cylinder is at rest relative to the Earth.

$t \approx 6.26$ s.

Since the person is stationary relative to the Earth, the forces applied to him are balanced. Thus, the tension force is numerically equal to the weight of the person $m_2 g$. Therefore, for the weight, Newton's second law will be written as: $m_1 a = m_2 g - m_1 g = (m_2 - m_1)g$. Hence the acceleration of the weight $a = g \frac{m_2 - m_1}{m_1}$ and is directed upwards. The time for the weight to rise is determined from the formula $h = at^2/2$: $t = \sqrt{2h/a} = \sqrt{\frac{2hm_1}{g(m_2 - m_1)}}.$

1. Both boys will reach the middle of the distance between them simultaneously after time $\tau = l/(3v)$; if the masses of the boys are unequal, the meeting place divides the distance between them inversely proportional to the masses of the boys.

2. The gymnasts will reach the block simultaneously after time $\tau = l/(3v)$.

Since the forces applied to the boys are equal, according to Newton's third law, the accelerations, and consequently, the velocities of the boys relative to the Earth are equal. The length of the rope between the boys decreases with speed $3v$ and the boys will reach the middle of the distance between them after time $\tau = l/(3v)$. If the masses of the boys are different, the accelerations and paths traveled by them relative to the Earth are inversely proportional to the masses of the boys.

• Compare the solution of this problem with the solution of problem item 1.

0.5 N; 1 N.

The magnitude of the static friction force cannot exceed $F_{fr.max} = kN$, where $N$ is the normal pressure. If the body is at rest, the friction force is equal in magnitude to the sum of the projections of all forces acting on the body onto the direction of possible motion (i.e., onto the direction in which the body would start moving if the friction force were zero). If the force acting on the body in the direction of possible motion exceeds $F_{fr.max}$, the body moves with acceleration, and the friction force is constant and equal to $F_{fr.max}$.

The graph is shown below.

The graph is shown below.

The motion of a body located on an inclined plane becomes possible if $mg \sin \alpha \ge kmg \cos \alpha$, i.e., at an angle $\alpha \ge \alpha_0 = \arctan k$. If $\alpha < \alpha_0$, the body is at rest and the friction force is equal to $mg \sin \alpha$ (see problem 23). If $\alpha \ge \alpha_0$, then $F_{\text{fr}} = F_{\text{fr.max}} = kmg \cos \alpha$.

$k = \tan \alpha$.

1. $a = (1 - m_1/m_2)g$; $F_{\text{fr}} = m_1 g$.

2. $a_1 = \dfrac{m_1 g - m_2(g-a_2)}{m_1+m_2}$; $F_{\text{fr}} = \dfrac{m_1 m_2 (2g - a_2)}{m_1+m_2}$.

1. Since weight $m_2$ is stationary, the tension of the string, and consequently, the friction force of the ring on the string are equal to $m_2 g$. [Note: Error in original text, should be $m_1 g$ based on context].

2. The friction force is equal to the tension force of the string, and the acceleration of the ring relative to the block is $a_2 - a_1$. The equations of motion for the weights are expressed as follows: $m_1 a_1 = m_1 g - F_{\text{fr}};$ $m_2(a_2 - a_1) = m_2 g - F_{\text{fr}}.$

$F = \dfrac{M(kg+a)}{\cos \alpha + k \sin \alpha}$.

The normal pressure force is $Mg - F \sin \alpha$ (see figure, a) or $Mg + F \sin \alpha$ (see figure, b) depending on the direction of action of force $\vec{F}$.

1. $T = g \frac{m_1 m_2 (1+k)}{m_1+m_2} = 1.47$ H. 2. Does not change. 3. $F = 2.08$ H.

$T = \frac{m_1 m_2 (g+a)(1+k)}{m_1+m_2}$ for $km_1 < m_2$; $T = m_2(g+a)$ for $km_1 > m_2$. The state of motion (rest) does not change.

The accelerations of the weights relative to the table are equal in magnitude (weights are connected by an inextensible string): $a'_1 = a'_2 = a'$. In a stationary reference frame, the acceleration of weight $m_2$ is directed vertically, and $a_2 = a' - a$. The acceleration of weight $m_1$ has vertical and horizontal components: $a_{1v} = a$, $a_{1h} = a'$. Applying Newton's second law, we get the system of equations: $m_2(a'-a) = m_2 g - T$, $m_1 a' = T - kN$, $m_1 a = N - m_1 g$, where $T$ is the string tension; $N$ is the reaction force of the table. Solving this system, we find the answer. Note. This system of equations can be represented as $m_2 a' = m_2(g+a) - T$, $m_1 a' = T - kN$, $N - m_1(g+a) = 0$. Thus, the upward movement of the elevator with acceleration $a$ is equivalent to a reference frame in which the acceleration of free fall is $g+a$. If $a'=0$ (for $km_1 > m_2$), the answer follows from the equation $m_2 a = T - m_2 g$. In this case, the friction force is no longer equal to $kN$ and the second equation is incorrect. From this note, the answer to the second question of the problem immediately follows.

$F_1 = 38.4$ H; $F_2 = 64$ H; does not change.

$F = (n+1)m(a+fg)$; $X_1 = m l \frac{a+fg}{k}$.

$a = \frac{F_1(\cos \alpha + k \sin \alpha) - F_2(\cos \beta - k \sin \beta)}{m_1+m_2} - gk$; $T = \frac{F_1 m_2 (\cos \alpha + k \sin \alpha) + F_2 m_1 (\cos \beta - k \sin \beta)}{m_1+m_2}$.

See problem 28.

1. $a_B = \dfrac{F(\cos \alpha + k_1 \sin \alpha)}{m_1} - k_1 g$; $a_A = \dfrac{(k_1-k_2)(m_1 g - F \sin \alpha)}{m_2} - k_2 g$.

2. $k_2 \ge \dfrac{m_1 k_1 + m_2 \tan \alpha}{m_1+m_2}$.

1. The pressure forces of block B on block A and block A on the table are respectively $m_1 g - F \sin \alpha$ and $(m_1+m_2)g - F \sin \alpha.$.

2. Since block B slides on the plate, two forces act on the plate from block B: the normal pressure force $N_1 = m_1 g \cos \alpha$ and the friction force $F_{fr1} = k_1 N_1 = k_1 m_1 g \cos \alpha.$ The normal pressure force of the plate on the inclined plane is $N_2 = N_1 + m_2 g \cos \alpha = (m_1+m_2)g \cos \alpha.$ The plate does not slide on the inclined plane if $m_2 g \sin \alpha + F_{fr1} \le F_{fr2} = k_2 N_2,$ i.e., if $m_2 g \sin \alpha + k_1 m_1 g \cos \alpha \le k_2 (m_1+m_2)g \cos \alpha$ or
   $k_2 \ge k_1 \frac{m_1}{m_1+m_2} + \frac{m_2}{m_1+m_2} \tan \alpha = \frac{m_1 k_1 + m_2 \tan \alpha}{m_1+m_2}$.

1) The acceleration of the body at all times during its motion is constant and equal to $g$;

2. In accordance with Newton's second law $a = (mg + F_{\text{res}})/m$. When moving upwards, the resistance force is directed, like gravity, downwards and decreases as the body ascends (since the speed of the body decreases), during descent, the resistance force is directed upwards and increases. Therefore, the acceleration of the body at the beginning of the motion is maximum (and greater than $g$), decreases during ascent and becomes equal to $g$ at the highest point of the trajectory, then continues to decrease during descent and may even become zero.

The speed of the large ball will be $\sqrt{2}$ times greater than the speed of the smaller ball.

Under steady motion, the air resistance force $F_{\text{res}}$ and the gravity force $Mg$ acting on the ball are equal, but $F_{\text{res}} = kv^2 (\pi D^2 / 4),$ and $M = \pi \rho D^3 / 6,$ where $\rho$ is the density of the ball; $D$ is its diameter. Equating $F_{\text{res}}$ and $Mg$, we find that $v \sim \sqrt{D}$.

$3g(m - \rho V)$.

See solutions to problems 35 and 36.

0.29 N.

a) $k=0.07$;

b) $a \approx 0.39$ m/s$^2$;

c) $t = 22.7$ s;

d) $v = 8.85$ m/s.

$t = \frac{1}{\sin \alpha} \sqrt{\frac{2h}{g(1 - \cot \alpha \tan \beta)}}$.

$T = (1/4) kmg \cos \alpha$.

On the upper weight along the inclined plane act the component of gravity $F = mg \sin \alpha$, the string tension force $T$, and the friction force $F_{\text{fr}} = kmg \cos \alpha$. According to Newton's second law, we get $ma = mg \sin \alpha - kmg \cos \alpha + T$. For the lower weight, we have accordingly $ma = mg \sin \alpha - (k/2)mg \cos \alpha - T$ (the weight moves with the same acceleration, and the coefficient of friction between it and the inclined plane is half as much). Subtracting the second from the first equation, we find the string tension force $T = (1/4) kmg \cos \alpha$.

$a = g \dfrac{m_2 - m_1(\sin \alpha + k \cos \alpha)}{m_1+m_2} \approx 0.42$ m/s$^2$; $T = g \dfrac{m_1 m_2}{m_1+m_2} (1 + \sin \alpha + k \cos \alpha) \approx 9.36$ H.

a) $a \approx 0.24$ m/s$^2$;

b) $T = 6$ H.

According to Newton's second law $m_1 g \sin \beta - km_1 g \cos \beta - T = m_1 a$; (1) $-m_2 g \sin \alpha - km_2 g \cos \alpha + T = m_2 a$. (2) Solving equations (1) and (2) together, we find $a$ and $T$.

$k = \tan \alpha \frac{n^2-1}{n^2+1} = 0.1$.

From the equations of motion of the stone up and down, find that its accelerations are respectively equal: $a_1 = g(\sin \alpha + k \cos \alpha)$; $a_2 = g(\sin \alpha - k \cos \alpha)$. (1) (2) The time of motion of the stone $t_1 = \sqrt{2l/a_1}$; this follows from two equations: $l = v_0 t_1 - a_1 t_1^2 / 2$, $v_0 - a_1 t_1 = 0$, and $t_2 = \sqrt{2l/a_2}$, where $l$ is the distance to the highest point of ascent of the stone. Hence $t_1/t_2 = \sqrt{a_2/a_1} = 1/n$ (according to the condition). (3) Solving equations (1), (2), and (3) together, find the value of $k$.

$a > kg = 0.1g$.

The force acting on the weight from the side of the board cannot be greater than the maximum static friction force.

$a > dg/h = 0.1g$.

1st method. The acceleration $a$ of the body is provided by the resultant of the force $F$ and the force of gravity $P=mg$ (see figure). The force $F$ is created by the normal pressure force $N$, applied in the limiting case (before tipping) at point C, and the friction force $F_{\text{fr}}$. If the acceleration is greater than the limiting value, the corresponding force $F$ cannot be provided, since for this the force $N$ (which is the resultant of the distributed pressure forces on the support) would have to be applied outside the support of the body. From the similarity of triangles OAB and COD $ma_{\text{lim}}/(mg) = d/h$, where $a_{\text{lim}}$ is the limiting acceleration at which the body does not yet tip over. Hence $a_{\text{lim}} = dg/h$; for tipping the body $a > a_{\text{lim}}$, i.e., $a > dg/h$.

2nd method. In the non-inertial reference frame associated with the sheet of paper, the force $F = -ma_{\text{lim}}$ (see solution to problem 17) and the force of gravity $P=mg$ act on the body at the moment of detachment. At $a_{lim}$, the sum of the moments of these forces relative to point C is zero, i.e., $-ma_{\text{lim}} h/2 + mg d/2 = 0$, hence $a_{\text{lim}} = dg/h$; for tipping $a > dg/h$.

$t = \sqrt{\dfrac{2l}{F/m - kg(1+m/M)}}$; $F_0 = kmg(1+m/M)$.

A force $F$ and a friction force $kmg$ act on the block. According to Newton's second law, its acceleration $a_1$ is determined by the equation $m a_1 = F - kmg$. A force $kmg$ acts on the cart from the side of the block, according to Newton's third law. Its acceleration $a_2$ is found from the equation $M a_2 = kmg$. The block slides relative to the cart with acceleration $a = a_1 - a_2 = F/m - kg(1+m/M)$. It travels the length $l$ of the cart in time $t = \sqrt{2l/a}$. The minimum value $F_0$ of force $F$ is determined by the condition $a=0$.

$t_1 = v_0 M / F$; $F_{fr1} = F$, $F_{fr2} = FM/(M+m)$; $l = l_0 + Mv_0^2 / (2F)$.

From the condition of constant velocity of the block relative to the stationary board, it follows that the friction force between them is maximum and equal to $F$. After cutting the cord, the block does not change its state of motion (sliding and, consequently, the equality to zero of the sum of forces acting on it is preserved). The board, however, begins to move with uniform acceleration under the action of the force $F_{\text{fr}1} = F$: $a_1 = F/M$ until some moment $t_1 = v_0/a_1 = Mv_0/F$. At moment $t_1$, the speed of the board equals the speed of the block, its sliding stops, and further they will move as a single unit with acceleration $a_2 = F/(M+m)$, and the static friction force $F_{fr2}$ will become equal to $F - ma_2 = FM/(M+m)$. The block travels a path $v_0 t_1 - a_1 t_1^2 / 2 = Mv_0^2 / (2F)$ along the board during time $t_1$. Thus, the minimum required length of the board, at which the block still manages to stop on it (neglecting the size of the block), is $l = l_0 + Mv_0^2 / (2F)$. Plot the velocity graph for both bodies.

$l = \dfrac{M}{M-m} s$.

After the wagon detaches, the train will move with acceleration imparted to it by a force equal to $F_{thrust} - F_{\text{res}} = F_{thrust} - (F_{res.init} - kmg) = kmg$ (the initial resistance $F_{res.init}$ was equal to the thrust force of the locomotive); $a_1 = kmg/(M-m)$. The wagon will move with uniform deceleration with acceleration $a_2 = -kg$. Relative to each other, the wagon and the locomotive move with acceleration $a_{rel} = a_1 - a_2 = \frac{kmg}{M-m} + kg = kg \frac{M}{M-m}$. The distance $l$ between them will relate to the path $s$ traveled by the wagon until stopping as $a_{rel}/|a_2|$. Another solution see in problem 40 in Chapter: Work, Power, Energy .