Spring Force Questions in English

(C) When a spring is pulled from both ends by forces of equal magnitude $F$,the tension in the spring is equal to the magnitude of the applied force $F$.
In this case,both ends of the spring are pulled by a force of $5 \, N$.
Therefore,the tension in the spring is $5 \, N$.

(C) The potential energy $U$ stored in a spring is given by the formula $U = \frac{F^2}{2k}$,where $F$ is the applied force and $k$ is the spring constant.
Since both springs are stretched with the same force $F$,the potential energy is inversely proportional to the spring constant: $U \propto \frac{1}{k}$.
Therefore,the ratio of the potential energies is $\frac{U_1}{U_2} = \frac{k_2}{k_1}$.
Given $k_1 = 1500 \ N/m$ and $k_2 = 3000 \ N/m$,we have:
$\frac{U_1}{U_2} = \frac{3000}{1500} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(C) First,calculate the force constant $(k)$ of the spring using Hooke's Law: $F = kx$,where $F = mg$.
Given: $m = 1.0\,kg$,$g = 10\,m/s^2$,and extension $x = 2\,cm = 0.02\,m$.
$k = \frac{mg}{x} = \frac{1.0 \times 10}{0.02} = 500\,N/m$.
Next,determine the total extension of the spring when its length becomes $60\,cm$. The original length is $50\,cm$,so the total extension $x_{total} = 60\,cm - 50\,cm = 10\,cm = 0.1\,m$.
The elastic potential energy $(U)$ stored in the spring is given by $U = \frac{1}{2}kx^2$.
$U = \frac{1}{2} \times 500 \times (0.1)^2 = 250 \times 0.01 = 2.5\,J$.

(A) According to Hooke's Law,the restoring force in a spring is given by $F = kx$,where $k$ is the spring constant and $x$ is the extension.
When a mass $m$ is hung from the spring,the force exerted is equal to the weight of the mass,$F = mg$.
Equating the two,we get $mg = kx$.
Therefore,the extension $x = \frac{mg}{k}$.
Given $m = 20\, kg$,$g = 9.8\, m/s^2$,and $k = 4000\, N/m$.
Substituting the values: $x = \frac{20 \times 9.8}{4000} = \frac{196}{4000} = 0.049\, m$.
To convert the extension into centimeters,multiply by $100$: $x = 0.049 \times 100 = 4.9\, cm$.

(A) The potential energy $U$ stored in a spring is given by the formula $U = \frac{1}{2}kx^2$.
According to Hooke's Law,the tension $T$ in the spring is related to the extension $x$ by the equation $T = kx$.
From this,we can express the extension as $x = \frac{T}{k}$.
Substituting this value of $x$ into the energy formula:
$U = \frac{1}{2}k \left( \frac{T}{k} \right)^2$
$U = \frac{1}{2}k \left( \frac{T^2}{k^2} \right)$
$U = \frac{T^2}{2k}$.
Therefore,the correct option is $A$.

(A) The force $F$ required to stretch a spring by a distance $x$ is given by $F = kx$,where $k$ is the spring constant.
For a spring of length $L$,the spring constant $k$ is inversely proportional to its length,i.e.,$k \propto \frac{1}{L}$.
If the length of the spring is halved $(L' = L/2)$,the new spring constant $k'$ becomes $k' = k \cdot (L/L') = k \cdot (L / (L/2)) = 2k$.
The slope of the force-displacement graph is equal to the spring constant $k$. Since the new spring constant $k'$ is double the original value,the slope of the line $OA$ increases.
An increase in the slope of a line passing through the origin in an $F-x$ graph means the line shifts towards the $F$-axis.
Therefore,the correct option is $A$.

(A) According to Hooke's law,the restoring force $F$ in a spring is directly proportional to the extension $l$ produced in it.
Mathematically,$F \propto l$.
Removing the proportionality sign,we introduce a spring constant $k$:
$F = kl$,where $k$ is the spring constant.

(D) The initial spring constant is $k$. When a spring of length $l$ is cut into two equal parts,the spring constant of each part becomes $k' = 2k$.
When these two parts are connected in parallel,the equivalent spring constant $K_{eq}$ is the sum of the individual constants:
$K_{eq} = k' + k' = 2k + 2k = 4k$.
According to Hooke's Law,$F = kx$. For the original spring,$W = kx$.
For the parallel combination,the same weight $W$ is applied,so $W = K_{eq} \times x'$,where $x'$ is the new extension.
Substituting the values: $kx = 4k \times x'$.
Solving for $x'$,we get $x' = \frac{kx}{4k} = \frac{x}{4}$.

(D) The force constant $k$ of a spring is inversely proportional to its length $l$,i.e.,$k \propto \frac{1}{l}$.
Let the total length of the spring be $L$. The lengths of the two pieces are $l_1 = \frac{1}{3}L$ and $l_2 = \frac{2}{3}L$.
Let $k_1$ and $k_2$ be the force constants of the shorter and longer pieces,respectively.
Since $k_1 l_1 = k_2 l_2 = kL$,we have $\frac{k_1}{k_2} = \frac{l_2}{l_1}$.
Substituting the values,$\frac{k_1}{k_2} = \frac{2/3 L}{1/3 L} = \frac{2}{1}$.
Therefore,the ratio of the force constants is $2:1$.

(B) The force constant $K$ of a spring is inversely proportional to its natural length $l$,i.e.,$K \propto \frac{1}{l}$ or $Kl = \text{constant}$.
When a spring of length $l$ is cut into two parts,the product of the force constant and the length of each part remains constant.
Initially,the spring has length $l$ and force constant $K$.
One-fourth of the length is cut away,so the remaining length of the spring is $l' = l - \frac{1}{4}l = \frac{3}{4}l$.
Let the new force constant be $K'$.
According to the relation $Kl = K'l'$,we have:
$K \cdot l = K' \cdot (\frac{3}{4}l)$
$K = K' \cdot \frac{3}{4}$
$K' = \frac{4}{3}K$.

(C) For springs connected in series,the effective force constant $k_{eff}$ is given by the formula: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \dots$
Substituting the given values: $\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{4k} + \frac{1}{8k} + \dots$
Factor out $\frac{1}{k}$: $\frac{1}{k_{eff}} = \frac{1}{k} \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right)$
The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.
The sum $S$ of an infinite geometric progression is given by $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
Therefore,$\frac{1}{k_{eff}} = \frac{1}{k} \times 2 = \frac{2}{k}$.
Thus,$k_{eff} = \frac{k}{2}$.

(B) Let $x$ be the maximum extension of the spring.
Since the block starts from rest and momentarily comes to rest at the point of maximum extension,the change in kinetic energy is zero.
According to the work-energy theorem,the total work done by all forces is zero.
The forces acting on the block are gravity ($Mg$ downwards) and the spring force ($-Kx$ upwards).
Work done by gravity = $Mgx$
Work done by spring = $-\frac{1}{2}Kx^2$
Equating the total work to zero:
$Mgx - \frac{1}{2}Kx^2 = 0$
$Mgx = \frac{1}{2}Kx^2$
Since $x \neq 0$,we can divide by $x$:
$Mg = \frac{1}{2}Kx$
$x = \frac{2Mg}{K}$

(A) The potential energy $U$ stored in a stretched spring is given by $U = \frac{1}{2} k x^2$.
According to Hooke's Law,the tension force $T$ in the spring is $T = kx$,which implies $x = \frac{T}{k}$.
Substituting the value of $x$ into the energy formula:
$U = \frac{1}{2} k \left( \frac{T}{k} \right)^2$
$U = \frac{1}{2} k \left( \frac{T^2}{k^2} \right)$
$U = \frac{T^2}{2k}$.

(A) Let $x_{max}$ be the maximum compression and $x_{eq}$ be the compression at equilibrium.
$(i)$ For maximum compression,the loss in gravitational potential energy of the block equals the gain in elastic potential energy of the spring:
$mgx_{max} = \frac{1}{2} k x_{max}^2$
$x_{max} = \frac{2mg}{k}$
$(ii)$ At the equilibrium position,the spring force balances the weight of the block:
$kx_{eq} = mg$
$x_{eq} = \frac{mg}{k}$

(A) Since the spring force is an internal force,the net external force on the system is zero.
Therefore,the acceleration of the center of mass is zero: $M_A \vec{a}_A + M_B \vec{a}_B = 0$.
Given: $M_A = 100 \ gm = 0.1 \ kg$,$M_B = 250 \ gm = 0.25 \ kg$.
Acceleration of ball $B$,$\vec{a}_B = 10 \ cm/s^2$ (West).
Using the equation: $\vec{a}_A = -\frac{M_B}{M_A} \vec{a}_B$.
$\vec{a}_A = -\frac{0.25 \ kg}{0.10 \ kg} \times (10 \ cm/s^2 \text{ West})$.
$\vec{a}_A = -2.5 \times (10 \ cm/s^2 \text{ West}) = -25 \ cm/s^2 \text{ West}$.
Since the negative sign indicates the opposite direction,$\vec{a}_A = 25 \ cm/s^2$ towards the east.

(B) When the body is released slowly,the spring reaches equilibrium where the spring force equals the gravitational force: $kx = mg$,which gives $x = \frac{mg}{k}$.
When the body is released suddenly,the potential energy lost by the body is stored as elastic potential energy in the spring at the maximum extension $y$. By the law of conservation of energy: $mgy = \frac{1}{2}ky^2$.
Solving for $y$: $y = \frac{2mg}{k}$.
Since $x = \frac{mg}{k}$,we substitute this into the equation for $y$ to get $y = 2x$.

(B) The force constant $k$ of a spring is inversely proportional to its length $l$, i.e., $k \propto 1/l$ or $kl = \text{constant}$.
Let the total length of the spring be $L$, so $kL = \text{constant}$.
Let the two pieces have lengths $l_1$ and $l_2$ such that $l_1 + l_2 = L$.
Given $l_1 = 2l_2$, therefore $L = 2l_2 + l_2 = 3l_2$.
For the larger piece (length $l_1 = 2l_2$), let the force constant be $k_1$.
Using $k_1 l_1 = kL$, we get $k_1 (2l_2) = k(3l_2)$.
Solving for $k_1$, we get $k_1 = \frac{3}{2}k$.

(D) For the original spring,the force constant is $k$ and the extension is $x = W/k$.
When a spring is cut into two equal parts,the force constant of each part becomes $k' = 2k$.
When these two parts are connected in parallel,the equivalent force constant $k_{eq}$ is given by $k_{eq} = k' + k' = 2k + 2k = 4k$.
The new extension $x'$ produced by the same weight $W$ is given by $x' = W / k_{eq}$.
Substituting $k_{eq} = 4k$,we get $x' = W / (4k)$.
Since $x = W/k$,we have $x' = x/4$.

(C) Before the string is cut,the system is in equilibrium.
For block $B$: The tension $T$ in the string is equal to the weight of block $B$,so $T = mg$.
For block $A$: The spring force $kx$ balances the weight of both blocks $A$ and $B$ plus the tension $T$. However,looking at the free body diagram of $A$,$kx = T + 3mg = mg + 3mg = 4mg$.
Immediately after the string is cut,the tension $T$ becomes zero,but the spring force $kx$ remains $4mg$ because the spring does not change its length instantaneously.
For block $B$: The only force acting on it is gravity,so its acceleration is $a_B = g$ (downwards).
For block $A$: The net force is $F_{net} = kx - 3mg = 4mg - 3mg = mg$ (upwards).
Therefore,the acceleration of $A$ is $a_A = \frac{F_{net}}{3m} = \frac{mg}{3m} = \frac{g}{3}$ (upwards).
Thus,the accelerations of $A$ and $B$ are $\frac{g}{3}$ and $g$ respectively.

(A) Let the maximum extension in the spring be $x$.
According to the law of conservation of energy,the loss in gravitational potential energy of the block is equal to the gain in elastic potential energy of the spring.
Loss in gravitational potential energy = $Mgx$.
Gain in elastic potential energy = $\frac{1}{2} k x^2$.
Equating the two: $Mgx = \frac{1}{2} k x^2$.
Solving for $x$ (where $x \neq 0$): $x = \frac{2Mg}{k}$.

(A) Let the original length of the spring be $L$. When the spring is cut into lengths in the ratio $1:2:3$,the lengths of the three segments are $l_1 = L/6$,$l_2 = 2L/6$,and $l_3 = 3L/6$.
Since the force constant $k$ is inversely proportional to the length $(k \propto 1/l)$,the new force constants are:
$k_1 = k(L/l_1) = 6k$
$k_2 = k(L/l_2) = 3k$
$k_3 = k(L/l_3) = 2k$
When connected in series,the equivalent force constant $k'$ is given by:
$1/k' = 1/k_1 + 1/k_2 + 1/k_3 = 1/(6k) + 1/(3k) + 1/(2k) = (1+2+3)/(6k) = 6/(6k) = 1/k$
Thus,$k' = k$.
When connected in parallel,the equivalent force constant $k''$ is given by:
$k'' = k_1 + k_2 + k_3 = 6k + 3k + 2k = 11k$.
Therefore,the ratio $k':k'' = k : 11k = 1:11$.

(B) The spring constant $k$ of a spring is inversely proportional to its length $l$,i.e.,$k \propto \frac{1}{l}$.
This is because the spring constant is given by $k = \frac{G J}{R^2 n d^3}$,where $n$ is the number of coils.
Since the number of coils $n$ is directly proportional to the length of the spring,we have $k \propto \frac{1}{n}$.
When the spring is cut into two equal halves,the number of coils in each new spring becomes $n' = \frac{n}{2} = \frac{10}{2} = 5$.
Therefore,the new spring constant $k'$ is given by $k' \times n' = k \times n$.
$k' \times \frac{n}{2} = k \times n$.
$k' = 2k$.

(B) For the $2 \text{ kg}$ mass,the net force is given by:
$F_{\text{net}} = 10 - T = m_1 a_1$
Substituting the values $m_1 = 2 \text{ kg}$ and $a_1 = 2 \text{ ms}^{-2}$:
$10 - T = 2 \times 2 = 4 \text{ N}$
$T = 10 - 4 = 6 \text{ N}$
For the $3 \text{ kg}$ mass,the spring force $T$ is the only horizontal force acting on it:
$F_{\text{net}} = T = m_2 a_2$
Substituting the values $T = 6 \text{ N}$ and $m_2 = 3 \text{ kg}$:
$6 = 3 \times a_2$
$a_2 = \frac{6}{3} = 2 \text{ ms}^{-2}$

(A) Let the spring force due to each spring be $F$.
Initially,the block is in equilibrium:
$2F = mg$
$2F = 3 \times 10 = 30 \; N$
$F = 15 \; N$
At $t = 0$,when spring $A$ is cut,the force in spring $B$ does not change instantaneously.
Applying Newton's second law for the block at $t = 0$:
$mg - F = ma$
$30 - 15 = 3a$
$15 = 3a$
$a = 5 \; m/s^2$

(B) Let $a$ be the initial downward acceleration of the upper block of mass $2m$.
Before the string breaks,the system is in equilibrium.
For the lower block of mass $m$,the spring force $F_s = kx = mg$ (upward).
For the upper block of mass $2m$,the tension in the string $T = 2mg + F_s = 2mg + mg = 3mg$.
Immediately after the string breaks,the tension $T$ becomes $0$,but the spring force $F_s$ remains $mg$ (as the spring length cannot change instantaneously).
Now,the forces acting on the upper block of mass $2m$ are its weight $2mg$ (downward) and the spring force $F_s = mg$ (downward).
Applying Newton's second law for the upper block: $F_{net} = (2m)a$.
$2mg + F_s = (2m)a$.
$2mg + mg = (2m)a$.
$3mg = 2ma$.
Therefore,$a = 3g/2$.

(B) Let the spring constant be $k$. When the block $A$ is pushed towards the wall by a distance $x$,the spring is compressed by $x$.
The spring force exerted on block $B$ is $F_s = kx$,directed towards the wall.
Since block $B$ is in equilibrium in the horizontal direction,the normal reaction $N$ from the wall must balance the spring force.
Therefore,$N = F_s = kx$.
This shows that the normal reaction $N$ is directly proportional to the compression $x$ $(N \propto x)$.
As $x$ increases,$N$ increases linearly starting from $N = 0$ when $x = 0$.
Thus,the graph is a straight line passing through the origin with a positive slope.

(B) The velocity of the block is maximum when its acceleration is zero.
At any compression $x$ of the spring,the forces acting on the block are the gravitational force $mg$ (downward) and the spring force $kx$ (upward).
The net force $F_{net} = mg - kx$.
For maximum velocity,the net force must be zero,so $mg - kx = 0$,which gives $x = mg/k$.
Given $m = 1 \ kg$,$g = 10 \ m/s^2$ (using $g = 10 \ m/s^2$ for standard calculation),and $k = 2 \ kg/cm = 200 \ N/m$.
Converting units: $k = 2 \ kg/cm = 200 \ N/m = 2 \ N/cm$.
Thus,$x = (1 \ kg \times 10 \ m/s^2) / (200 \ N/m) = 10/200 \ m = 0.05 \ m = 5 \ cm$.
The height $h$ from the table surface is the original length of the spring minus the compression $x$.
$h = L - x = 20 \ cm - 5 \ cm = 15 \ cm$.

(D) For the rod $AB$ to remain horizontal,the extension $x$ in both springs must be equal. Let the extension be $x$.
The forces exerted by the springs at ends $A$ and $B$ are $F_A = k_1 x$ and $F_B = k_2 x$ respectively.
For rotational equilibrium about point $C$,the sum of moments must be zero:
$\sum \tau_C = 0$
$(k_1 x) \cdot AC = (k_2 x) \cdot BC$
Canceling $x$ from both sides:
$k_1 \cdot AC = k_2 \cdot BC$
$\frac{AC}{BC} = \frac{k_2}{k_1} \dots (i)$
We know that $AC + BC = l$,so $BC = l - AC$. Substituting this into equation $(i)$:
$\frac{AC}{l - AC} = \frac{k_2}{k_1}$
$k_1 \cdot AC = k_2 (l - AC)$
$k_1 \cdot AC = k_2 l - k_2 \cdot AC$
$AC (k_1 + k_2) = k_2 l$
$AC = \frac{l k_2}{k_1 + k_2}$

(A) The stiffness $k$ of a spring is inversely proportional to its length $l$,i.e.,$k \propto \frac{1}{l}$.
Let the total length of the spring be $L = l_{A} + l_{B}$.
Given the ratio $l_{A} : l_{B} = 2 : 3$,we can write $l_{A} = \frac{2}{5}L$ and $l_{B} = \frac{3}{5}L$.
Since $k \cdot l = \text{constant}$,we have $k_{A} \cdot l_{A} = k \cdot L$.
Substituting $l_{A} = \frac{2}{5}L$ into the equation:
$k_{A} \cdot (\frac{2}{5}L) = k \cdot L$
$k_{A} = k \cdot \frac{L}{\frac{2}{5}L} = \frac{5}{2}k$.

(C) For the $5\ kg$ block,the only horizontal force acting is the spring force $F_s = kx$. According to Newton's second law,$F_s = m_1 a_1$.
$F_s = 5\ kg \times 2\ m/s^2 = 10\ N$.
For the $15\ kg$ block,the forces acting are the applied force $F = 100\ N$ to the right and the spring force $F_s = 10\ N$ to the left.
According to Newton's second law,$F - F_s = m_2 a_2$.
$100\ N - 10\ N = 15\ kg \times a_2$.
$90\ N = 15\ kg \times a_2$.
$a_2 = \frac{90}{15} = 6\ m/s^2$.

(B) Case $1$: When the block is lowered very gradually,the acceleration is zero at all times. The spring force balances the weight of the block at equilibrium.
$mg = kd$
$\therefore d = \frac{mg}{k}$
Case $2$: When the block is released suddenly from the natural length of the spring,the block starts from rest and reaches its maximum expansion $x_{max}$ where its velocity again becomes zero.
Applying the Work-Energy Theorem between the initial position (natural length) and the final position (maximum extension):
$W_{gravity} + W_{spring} = \Delta K$
$mgx_{max} - \frac{1}{2}kx_{max}^2 = 0 - 0$
$mgx_{max} = \frac{1}{2}kx_{max}^2$
$x_{max} = \frac{2mg}{k}$
Since $d = \frac{mg}{k}$,we have $x_{max} = 2d$.

(C) Let $T_1$ be the tension in the right string and $F_s$ be the spring force in the left string.
At point $O$,the forces are in equilibrium.
Resolving forces horizontally: $F_s \cos 37^\circ = T_1 \cos 53^\circ$.
Since $\cos 37^\circ = 4/5$ and $\cos 53^\circ = 3/5$,we have $F_s (4/5) = T_1 (3/5) \implies 4F_s = 3T_1 \implies T_1 = \frac{4}{3}F_s$.
Resolving forces vertically: $F_s \sin 37^\circ + T_1 \sin 53^\circ = 200$.
Substituting $T_1$: $F_s (3/5) + (\frac{4}{3}F_s) (4/5) = 200$.
$\frac{3}{5}F_s + \frac{16}{15}F_s = 200 \implies \frac{9+16}{15}F_s = 200 \implies \frac{25}{15}F_s = 200$.
$F_s = 200 \times \frac{15}{25} = 200 \times 0.6 = 120\, N$.
Using Hooke's Law $F_s = kx$,where $x = 4\, cm = 0.04\, m$.
$120 = k \times 0.04 \implies k = \frac{120}{0.04} = 3000\, N/m$.

(C) The force constant $K$ of a spring is inversely proportional to its length $\ell$,given by the relation $K \propto \frac{1}{\ell}$.
When the spring is cut into half,the new length becomes $\ell' = \frac{\ell}{2}$.
Let the new force constant be $K'$.
Using the proportionality,we have $\frac{K'}{K} = \frac{\ell}{\ell'} = \frac{\ell}{\ell / 2} = 2$.
Therefore,the new force constant is $K' = 2K$.

(B) By the Law of Conservation of Mechanical Energy $(COME)$,the total initial mechanical energy is equal to the total final mechanical energy.
Initial kinetic energy of the system is $KE_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
At the point of maximum elongation $x$,the relative velocity between the two blocks becomes zero,meaning both blocks move with the same velocity (which is zero in the center of mass frame).
Final potential energy of the spring is $PE_f = \frac{1}{2}kx^2$.
Applying conservation of energy: $KE_i = PE_f$.
$mv^2 = \frac{1}{2}kx^2$.
$x^2 = \frac{2mv^2}{k}$.
$x = \sqrt{\frac{2mv^2}{k}}$.

(B) Let the acceleration of the $1 \, kg$ mass be $a_1 = 10 \, m/s^2$ (towards the right).
Let the acceleration of the $2 \, kg$ mass be $a_2$.
Applying Newton's second law for the system of two masses connected by a spring,the external force $F_{ext}$ acting on the system is equal to the sum of the products of mass and acceleration of each individual body:
$F_{ext} = m_1 a_1 + m_2 a_2$
Given $F_{ext} = 20 \, N$,$m_1 = 1 \, kg$,$m_2 = 2 \, kg$,and $a_1 = 10 \, m/s^2$:
$20 = (1 \times 10) + (2 \times a_2)$
$20 = 10 + 2 a_2$
$10 = 2 a_2$
$a_2 = 5 \, m/s^2$
Thus,the acceleration of the $2 \, kg$ mass is $5 \, m/s^2$.

(A) Let the original length of the spring be $l$ and the spring constant be $k$. When the body is whirled in a horizontal circle,the centripetal force is provided by the spring force $F = kx$,where $x$ is the elongation.
The radius of the circular path is $r = l + x$.
The centripetal force is $F = m r \omega^2 = m(l + x) \omega^2$.
Case $1$: Elongation $x_1 = 1\, cm$,angular velocity $\omega_1 = \omega$.
$k(1) = m(l + 1) \omega^2$ --- $(i)$
Case $2$: Elongation $x_2 = 5\, cm$,angular velocity $\omega_2 = 2\omega$.
$k(5) = m(l + 5) (2\omega)^2 = 4m(l + 5) \omega^2$ --- $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{5k}{k} = \frac{4m(l + 5) \omega^2}{m(l + 1) \omega^2}$
$5 = \frac{4(l + 5)}{l + 1}$
$5(l + 1) = 4(l + 5)$
$5l + 5 = 4l + 20$
$l = 15\, cm$.

(D) The force constant $k$ of a spring is inversely proportional to its unstretched length,i.e.,$k \propto \frac{1}{\ell}$,which implies $k\ell = C$ (a constant).
For the two pieces,we have $k_1 \ell_1 = C$ and $k_2 \ell_2 = C$.
Therefore,$k_1 \ell_1 = k_2 \ell_2$.
Rearranging for the ratio,we get $\frac{k_1}{k_2} = \frac{\ell_2}{\ell_1}$.
Given that $\ell_1 = n\ell_2$,we substitute this into the ratio:
$\frac{k_1}{k_2} = \frac{\ell_2}{n\ell_2} = \frac{1}{n}$.

(A) The potential energy $E$ stored in a spring with spring constant $K$ when stretched by a force $F$ is given by the formula $E = \frac{F^2}{2K}$.
Since the applied force $F$ is the same for both springs,the energy stored is inversely proportional to the spring constant: $E \propto \frac{1}{K}$.
Therefore,the ratio of energies is $\frac{E_{A}}{E_{B}} = \frac{K_{B}}{K_{A}}$.
Given that $K_{A} = 2 K_{B}$,we have $\frac{K_{B}}{K_{A}} = \frac{1}{2}$.
Substituting this into the ratio,we get $\frac{E_{A}}{E_{B}} = \frac{1}{2}$.
Thus,$E_{B} = 2 E_{A}$.

(D) The force constant $k$ of a spring is inversely proportional to its length $l$,given by the relation $k \propto 1/l$ or $kl = \text{constant}$.
When a spring of length $l$ and force constant $k$ is cut into two equal halves,the length of each new piece becomes $l' = l/2$.
Since $k'l' = kl$,we have $k'(l/2) = kl$.
Solving for $k'$,we get $k' = 2k$.
Therefore,the force constant of each half is $2k$.

(A) For the $2\,kg$ mass,the forces acting are the applied force of $10\,N$ to the right and the spring force $F_s = kx$ to the left.
Applying Newton's second law for the $2\,kg$ mass:
$F_{net} = ma$
$10 - kx = 2 \times 2$
$10 - kx = 4$
$kx = 6\,N$
Now,for the $3\,kg$ mass,the only horizontal force acting on it is the spring force $kx$ pulling it to the right.
Applying Newton's second law for the $3\,kg$ mass:
$F_{net} = ma$
$kx = 3 \times a_{3kg}$
$6 = 3 \times a_{3kg}$
$a_{3kg} = \frac{6}{3} = 2\,ms^{-2}$

(D) When a spring of force constant $k$ is cut into $n$ equal pieces,the force constant of each piece becomes $k' = nk$.
Here,the spring is cut into $3$ equal pieces,so the force constant of each piece is $k' = 3k$.
When these $3$ pieces are connected in parallel,the equivalent force constant $k_{eq}$ is the sum of the individual force constants:
$k_{eq} = k' + k' + k' = 3k' = 3(3k) = 9k$.

(B) Let $m_1 = 10\, kg$ and $m_2 = 20\, kg$. Let $f$ be the spring force acting on both masses.
For the $10\, kg$ mass,the equation of motion is: $f = m_1 a_1$,where $a_1 = 12\, m\, s^{-2}$.
Thus,$f = 10 \times 12 = 120\, N$.
For the $20\, kg$ mass,the external force $F = 200\, N$ acts in the direction of motion,and the spring force $f$ acts in the opposite direction.
The equation of motion is: $F - f = m_2 a_2$.
Substituting the values: $200 - 120 = 20 \times a_2$.
$80 = 20 \times a_2$.
$a_2 = \frac{80}{20} = 4\, m\, s^{-2}$.

(A) Let the natural length of the spring be $\ell_{0}$ and the spring constant be $k$.
According to Hooke's Law,$F = k(\text{extension}) = k(\ell - \ell_{0})$.
For $F = 4\,N$,the length is $\alpha$: $4 = k(\alpha - \ell_{0}) \dots(i)$
For $F = 5\,N$,the length is $\beta$: $5 = k(\beta - \ell_{0}) \dots(ii)$
Subtracting $(i)$ from $(ii)$: $5 - 4 = k(\beta - \ell_{0}) - k(\alpha - \ell_{0}) \Rightarrow 1 = k(\beta - \alpha) \Rightarrow k = \frac{1}{\beta - \alpha}$.
Substitute $k$ into $(i)$: $4 = \frac{1}{\beta - \alpha}(\alpha - \ell_{0}) \Rightarrow 4(\beta - \alpha) = \alpha - \ell_{0} \Rightarrow \ell_{0} = \alpha - 4\beta + 4\alpha = 5\alpha - 4\beta$.
For $F = 9\,N$,let the length be $\gamma$: $9 = k(\gamma - \ell_{0}) \Rightarrow \gamma = \ell_{0} + \frac{9}{k}$.
Substituting $\ell_{0}$ and $k$: $\gamma = (5\alpha - 4\beta) + 9(\beta - \alpha) = 5\alpha - 4\beta + 9\beta - 9\alpha = 5\beta - 4\alpha$.

(B) Let $l$ be the original length of the spring and $k$ be the spring constant.
When the body is whirled in a horizontal plane,the centripetal force is provided by the spring force.
For the first case,the angular velocity is $\omega$ and elongation $x_1 = 1 \, cm$:
$m \omega^2 (l + x_1) = k x_1$ --- $(1)$
For the second case,the angular velocity is $2\omega$ and elongation $x_2 = 5 \, cm$:
$m (2\omega)^2 (l + x_2) = k x_2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{4 \omega^2 (l + x_2)}{\omega^2 (l + x_1)} = \frac{k x_2}{k x_1}$
$4 \frac{l + 5}{l + 1} = \frac{5}{1}$
$4(l + 5) = 5(l + 1)$
$4l + 20 = 5l + 5$
$l = 15 \, cm$.
Thus,the original length of the spring is $15 \, cm$.

(C) The force constant $k$ of a spring is inversely proportional to its length $l$,i.e.,$k \propto \frac{1}{l}$ or $kl = \text{constant}$.
Let the total length of the spring be $L$ and its force constant be $k$. The spring is cut into two pieces of lengths $l_1$ and $l_2$ such that $l_2 = 3l_1$.
The total length is $L = l_1 + l_2 = l_1 + 3l_1 = 4l_1$.
For the original spring,$kL = C$,where $C$ is a constant.
For the longer piece of length $l_2$,let the new force constant be $k_2$. Then $k_2 l_2 = C$.
Equating the two,$kL = k_2 l_2$.
Substituting the values,$k(4l_1) = k_2(3l_1)$.
Solving for $k_2$,we get $k_2 = \frac{4k}{3}$.

(C) The energy stored in a spring stretched by a force $F$ is given by $U = \frac{F^2}{2k}$.
Given that the forces applied to both springs are equal,let $F_P = F_Q = F$.
The energy stored in spring $P$ is $U_P = \frac{F^2}{2k_P}$.
The energy stored in spring $Q$ is $U_Q = \frac{F^2}{2k_Q} = E$.
Since $k_Q = \frac{k_P}{2}$,we have $k_P = 2k_Q$.
Substituting this into the expression for $U_P$:
$U_P = \frac{F^2}{2(2k_Q)} = \frac{1}{2} \left( \frac{F^2}{2k_Q} \right) = \frac{1}{2} E$.
Therefore,the energy stored in spring $P$ is $E/2$.

(N/A) $1$. Example of a variable force: The gravitational force between two objects as their distance changes,or the spring force $F = -kx$ where the force depends on the displacement $x$.
$2$. Formula for Hooke's law: $F = -kx$,where $F$ is the restoring force,$k$ is the spring constant,and $x$ is the displacement from the equilibrium position.

(N/A) The spring constant,denoted by $k$,is a measure of the stiffness of a spring. It is defined as the restoring force per unit extension or compression of the spring.
According to Hooke's Law,the restoring force $F$ is given by $F = -kx$,where $x$ is the displacement from the equilibrium position.
Therefore,the spring constant is $k = |F/x|$.
The $SI$ unit of the spring constant is Newtons per meter,denoted as $N/m$.

(A) The force applied is $F = 20\,dyne$.
The extension in the spring is $\Delta l = 1\,mm = 0.1\,cm$.
The force constant $k$ is given by Hooke's Law: $F = k \Delta l$.
Therefore,$k = \frac{F}{\Delta l} = \frac{20\,dyne}{0.1\,cm} = 200\,dyne/cm$.

(A) The force required to stretch a spring is given by Hooke's Law: $F = k \Delta x$.
Given:
Force constant $k = 0.5 \, Nm^{-1}$.
Extension $\Delta x = 10 \, cm = 10 \times 10^{-2} \, m = 0.1 \, m$.
Substituting the values into the formula:
$F = 0.5 \times 0.1 = 0.05 \, N$.
Therefore,the force necessary is $0.05 \, N$.