Work Done by Spring and Potential Energy of Spring Questions in English

(C) The work done $W$ in stretching a spring from an initial extension $x_1$ to a final extension $x_2$ is given by the formula:
$W = \frac{1}{2}k(x_2^2 - x_1^2)$
Given:
Force constant $k = 5 \times 10^3 \ N/m$
Initial extension $x_1 = 5 \ cm = 5 \times 10^{-2} \ m$
Final extension $x_2 = 10 \ cm = 10 \times 10^{-2} \ m$
Substituting the values:
$W = \frac{1}{2} \times (5 \times 10^3) \times [(10 \times 10^{-2})^2 - (5 \times 10^{-2})^2]$
$W = \frac{1}{2} \times 5000 \times [100 \times 10^{-4} - 25 \times 10^{-4}]$
$W = 2500 \times 75 \times 10^{-4}$
$W = 2500 \times 0.0075 = 18.75 \ J$ (or $N-m$).

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

(D) The potential energy $(U)$ stored in a spring stretched by a distance $(x)$ is given by the formula: $U = \frac{1}{2} k x^2$,where $k$ is the force constant of the spring.
From this relation,we can see that $U \propto x^2$.
Given that for an extension $x_1 = 2 \ cm$,the potential energy is $U_1 = U$.
For an extension $x_2 = 8 \ cm$,let the potential energy be $U_2$.
Using the ratio: $\frac{U_2}{U_1} = \left( \frac{x_2}{x_1} \right)^2$.
Substituting the values: $\frac{U_2}{U} = \left( \frac{8 \ cm}{2 \ cm} \right)^2 = (4)^2 = 16$.
Therefore,$U_2 = 16U$.

(C) Case $(a)$: When stretched by the same amount $x$,the work done is $W = \frac{1}{2} K x^2$. Since $K_P > K_Q$ and $x$ is the same,$W_P = \frac{1}{2} K_P x^2$ and $W_Q = \frac{1}{2} K_Q x^2$. Therefore,$W_P > W_Q$.
Case $(b)$: When stretched by the same force $F$,the work done is $W = \frac{F^2}{2K}$. Since $K_P > K_Q$ and $F$ is the same,$W_P = \frac{F^2}{2K_P}$ and $W_Q = \frac{F^2}{2K_Q}$. Since $K_P > K_Q$,it follows that $\frac{1}{K_P} < \frac{1}{K_Q}$,therefore $W_P < W_Q$ or $W_Q > W_P$.

(A) Given: Spring stiffness $K = 10\, N/cm = 1000\, N/m$.
Initial extension $x_1 = 4\, cm = 0.04\, m$.
Final extension $x_2 = 6\, cm = 0.06\, m$.
The work done $W$ to stretch a spring is given by the change in elastic potential energy:
$W = \frac{1}{2} K (x_2^2 - x_1^2)$
$W = \frac{1}{2} \times 1000 \times ((0.06)^2 - (0.04)^2)$
$W = 500 \times (0.0036 - 0.0016)$
$W = 500 \times 0.0020$
$W = 1.0\, J$.

(D) The initial potential energy of the spring is given by the formula $U = \frac{1}{2} k x^2$.
Given $U_1 = 10\, J$ for $x_1 = 0.3\, m$,we have $10 = \frac{1}{2} k (0.3)^2$.
Solving for the spring constant $k$: $k = \frac{20}{0.09} = \frac{2000}{9}\, N/m$.
The spring is stretched an additional $0.15\, m$,so the new total extension is $x_2 = 0.3 + 0.15 = 0.45\, m$.
The final potential energy is $U_2 = \frac{1}{2} k x_2^2 = \frac{1}{2} \times \frac{2000}{9} \times (0.45)^2$.
$U_2 = \frac{1000}{9} \times 0.2025 = 1000 \times 0.0225 = 22.5\, J$.
The work done to stretch the spring further is the change in potential energy: $W = U_2 - U_1 = 22.5\, J - 10\, J = 12.5\, J$.

(B) According to the work-energy theorem,the work done by all forces acting on a body is equal to the change in its kinetic energy.
For observer $A$ (ground frame),the initial kinetic energy of the block is $K_i = \frac{1}{2}mv_0^2$ and the final kinetic energy is $K_f = 0$ (since the block comes to rest).
The only force doing work on the block in the horizontal direction is the spring force $(F_s)$.
Therefore,$W_{spring} = K_f - K_i = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$.
Thus,the work done by the spring force as observed by $A$ is $-\frac{1}{2}mv_0^2$.

(B) Since the wedge is fixed,we can apply the principle of conservation of mechanical energy for the block of mass $M$.
Let $x_m$ be the maximum elongation in the spring.
At maximum elongation,the velocity of the block is zero.
The loss in gravitational potential energy of the block is equal to the gain in elastic potential energy of the spring.
$M g (x_m \sin \theta) = \frac{1}{2} k x_m^2$
Solving for $x_m$:
$x_m = \frac{2 M g \sin \theta}{k}$
Given $\theta = 37^\circ$,we know $\sin 37^\circ = \frac{3}{5}$.
Substituting the value of $\sin 37^\circ$:
$x_m = \frac{2 M g (3/5)}{k} = \frac{6 M g}{5 k}$

(B) The work done in stretching a spring by force $F$ is given by $W = \frac{F^2}{2k}$.
Given that for the same force $F$,$W_1 > W_2$,we have $\frac{F^2}{2k_1} > \frac{F^2}{2k_2}$,which implies $k_1 < k_2$. Thus,Statement $2$ is true.
When the springs are stretched by the same extension $x$,the work done is $W = \frac{1}{2}kx^2$.
Since $k_1 < k_2$,for the same extension $x$,$W_1 = \frac{1}{2}k_1x^2$ and $W_2 = \frac{1}{2}k_2x^2$. Therefore,$W_1 < W_2$.
Statement $1$ claims $W_1 > W_2$ for the same extension,which is false.
Thus,Statement $1$ is false and Statement $2$ is true.

(C) The potential energy stored in the compressed spring is converted into the kinetic energy of the particles in the center of mass frame.
Let the relative velocity of separation be $v_{rel}$.
The reduced mass of the system is $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{16 \times 2}{16 + 2} = \frac{32}{18} = \frac{16}{9} \ kg$.
The energy conservation equation is $\frac{1}{2} kx^2 = \frac{1}{2} \mu v_{rel}^2$.
Given $k = 100 \ Nm^{-1}$ and $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.
Substituting the values: $\frac{1}{2} \times 100 \times (\frac{1}{4})^2 = \frac{1}{2} \times \frac{16}{9} \times v_{rel}^2$.
$100 \times \frac{1}{16} = \frac{16}{9} \times v_{rel}^2$.
$v_{rel}^2 = \frac{100}{16} \times \frac{9}{16} = \frac{900}{256}$.
$v_{rel} = \sqrt{\frac{900}{256}} = \frac{30}{16} = 1.875 \ ms^{-1}$.
Rounding to two decimal places,we get $v_{rel} \approx 1.88 \ ms^{-1}$.

(B) At the equilibrium position,the spring force balances the gravitational force: $kx = mg$.
Given the force constant $k = \frac{2mg}{a}$,we substitute this into the equilibrium equation:
$\frac{2mg}{a} \cdot x = mg$
Solving for the displacement $x$:
$x = \frac{a}{2}$.
The work done by the Earth (gravitational force) on the block is given by $W = F \cdot d = mgx$.
Substituting the value of $x$:
$W = mg \left( \frac{a}{2} \right) = \frac{mga}{2}$.

(B) The potential energy stored in the spring when it is compressed by an amount $x$ is given by $U = \frac{1}{2}kx^2$.
Here,the natural length is $L_0$ and it is compressed to $L_0/2$,so the compression $x = L_0 - L_0/2 = L_0/2$.
By the principle of conservation of mechanical energy,the potential energy stored in the spring is converted into the kinetic energy of the block when the spring returns to its natural length.
$\frac{1}{2}mv^2 = \frac{1}{2}k\left(\frac{L_0}{2}\right)^2$
$mv^2 = k\frac{L_0^2}{4}$
$v^2 = \frac{k}{m} \cdot \frac{L_0^2}{4}$
$v = \sqrt{\frac{k}{m}} \cdot \frac{L_0}{2}$

(D) $1$. For the jumper to come to rest at distance $S$,the loss in gravitational potential energy equals the gain in elastic potential energy: $mgS = \frac{1}{2}K(S-l)^2$. Solving this quadratic equation $KS^2 - 2(Kl+mg)S + Kl^2 = 0$ gives $S = \frac{(Kl+mg) + \sqrt{2mgKl + m^2g^2}}{K}$. Thus,option $A$ is correct.
$2$. Maximum speed occurs at the equilibrium position where the net force is zero: $mg = Kx_{eq} \Rightarrow x_{eq} = \frac{mg}{K}$. The total distance fallen is $l + \frac{mg}{K}$. By conservation of energy: $mg(l + \frac{mg}{K}) = \frac{1}{2}mv^2 + \frac{1}{2}K(\frac{mg}{K})^2$. Solving for $v$ gives $v = \sqrt{2gl + \frac{mg^2}{K}}$. Thus,option $B$ is correct.
$3$. The time of free fall for distance $l$ is $t_1 = \sqrt{\frac{2l}{g}}$. Thus,option $C$ is correct.
$4$. The motion after the rope becomes taut is a simple harmonic motion starting from the natural length position with an initial downward velocity $v_0 = \sqrt{2gl}$. The time to reach the lowest point is not simply $\frac{\pi}{2}\sqrt{\frac{m}{K}} + \sqrt{\frac{2l}{g}}$. Therefore,option $D$ is the incorrect statement.

(B) Given: Mass $m = 1\, kg$,initial velocity $v_1 = 8\, m/s$,compression $x = 3\, m$.
Initial momentum $P_1 = m v_1 = 1 \times 8 = 8\, kg\cdot m/s$.
Final momentum $P_2 = P_1 / 2 = 4\, kg\cdot m/s$.
Since $P_2 = m v_2$,we have $v_2 = P_2 / m = 4 / 1 = 4\, m/s$.
By the law of conservation of energy,the loss in kinetic energy is equal to the potential energy stored in the spring:
$\frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2 = \frac{1}{2} k x^2$
$m(v_1^2 - v_2^2) = k x^2$
$1 \times (8^2 - 4^2) = k \times 3^2$
$1 \times (64 - 16) = 9k$
$48 = 9k$
$k = 48 / 9 = 16 / 3\, N/m$.

(C) The work done $W$ in stretching a spring by a force $F$ is given by the formula $W = \frac{1}{2} k x^2$.
Since the restoring force is $F = kx$,we have $x = \frac{F}{k}$.
Substituting $x$ into the work equation: $W = \frac{1}{2} k \left(\frac{F}{k}\right)^2 = \frac{F^2}{2k}$.
For a constant force $F$,the work done is inversely proportional to the force constant $k$ $(W \propto \frac{1}{k})$.
Given $k_1 > k_2$,it follows that $W_1 < W_2$.
Therefore,more work is done in the case of the second spring $(k_2)$.

(C) According to the Work-Energy Theorem,the work done by the spring force is equal to the change in kinetic energy of the block.
The work done by the spring force is $W_s = -\frac{1}{2}kx^2$.
The change in kinetic energy is $\Delta K = K_f - K_i = \frac{1}{2}m(v_f^2 - v_i^2)$.
Given the final speed $v_f = \frac{v}{4}$ and initial speed $v_i = v$,we have:
$-\frac{1}{2}kx^2 = \frac{1}{2}m\left(\left(\frac{v}{4}\right)^2 - v^2\right)$
$-\frac{1}{2}kx^2 = \frac{1}{2}m\left(\frac{v^2}{16} - v^2\right)$
$-\frac{1}{2}kx^2 = \frac{1}{2}m\left(-\frac{15v^2}{16}\right)$
$kx^2 = \frac{15mv^2}{16}$
$k = \frac{15mv^2}{16x^2}$

(C) At the time of maximum compression,both masses will have the same velocity $v_0$.
From the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_0$
$2 \times 10 + 5 \times 3 = (2 + 5) v_0$
$20 + 15 = 7 v_0$
$35 = 7 v_0 \Rightarrow v_0 = 5\,m/s$
From the law of conservation of energy,the initial kinetic energy equals the final kinetic energy plus the potential energy stored in the spring:
$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (m_1 + m_2) v_0^2 + \frac{1}{2} K x^2$
$\frac{1}{2} (2) (10)^2 + \frac{1}{2} (5) (3)^2 = \frac{1}{2} (2 + 5) (5)^2 + \frac{1}{2} (1120) x^2$
$100 + 22.5 = 87.5 + 560 x^2$
$122.5 = 87.5 + 560 x^2$
$35 = 560 x^2$
$x^2 = \frac{35}{560} = \frac{1}{16}$
$x = \sqrt{\frac{1}{16}} = 0.25\,m$.

(B) The work done on a spring stretched by force $F$ is $W = \frac{F^2}{2k}$. Since $W_1 > W_2$ for the same force $F$,we have $\frac{F^2}{2k_1} > \frac{F^2}{2k_2}$,which implies $k_1 < k_2$. Thus,Statement $-2$ is true.
When stretched by the same displacement $x$,the work done is $W = \frac{1}{2}kx^2$. Since $k_1 < k_2$,it follows that $\frac{1}{2}k_1x^2 < \frac{1}{2}k_2x^2$. Therefore,the work done on $S_1$ is less than that on $S_2$. Thus,Statement $-1$ is false.

(A) Let the vertical distance from the rod to the fixed end of the spring be $h$. From the geometry at position $B$,we have $\cos 53^{\circ} = \frac{h}{l_0+x}$ and $h = l_0$.
Thus,$\frac{3}{5} = \frac{l_0}{l_0+x}$.
Solving for $x$,we get $3(l_0+x) = 5l_0$,which implies $3x = 2l_0$,or $x = \frac{2}{3}l_0$.
By the law of conservation of mechanical energy,the potential energy stored in the spring at position $B$ is converted into the kinetic energy of the ring at position $A$ (where the spring length is $l_0$ and potential energy is zero).
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$.
$v = x\sqrt{\frac{k}{m}} = \left(\frac{2}{3}l_0\right)\sqrt{\frac{k}{m}}$.

(C) According to the Law of Conservation of Mechanical Energy $(COME)$,the total mechanical energy of the system remains constant.
Let the initial position of the ball be the reference level for gravitational potential energy $(PE = 0)$.
Initial state: The ball is at height $h$ above the platform. The spring is at its natural length. Initial kinetic energy $(KE_i)$ is $0$,and initial potential energy $(PE_i)$ is $0$ (if we measure from the release point).
Final state: The ball and the platform have moved down by a distance $x$. The ball is now at a vertical position $- (h + x)$ relative to the release point. The spring is compressed by $x$. Final kinetic energy $(KE_f)$ is $0$ (at the point of maximum compression).
Applying the principle of conservation of energy:
$PE_i + KE_i = PE_f + KE_f$
$0 + 0 = -mg(h + x) + \frac{1}{2}kx^2$
Rearranging the equation to solve for the spring constant $k$:
$mg(h + x) = \frac{1}{2}kx^2$
$k = \frac{2mg(h + x)}{x^2}$

(A) According to the work-energy theorem,the work done by the spring force is equal to the change in kinetic energy of the block.
The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.
The final kinetic energy when the speed is halved $(v' = v/2)$ is $K_f = \frac{1}{2}m(v/2)^2 = \frac{1}{8}mv^2$.
The work done by the spring force is $W = -\Delta PE = -\frac{1}{2}kx^2$.
By the work-energy theorem,$W = K_f - K_i$.
$-\frac{1}{2}kx^2 = \frac{1}{8}mv^2 - \frac{1}{2}mv^2$.
$-\frac{1}{2}kx^2 = \frac{mv^2 - 4mv^2}{8} = -\frac{3mv^2}{8}$.
$\frac{1}{2}kx^2 = \frac{3mv^2}{8}$.
$k = \frac{3mv^2}{4x^2}$.

(C) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by the formula $U = \frac{1}{2} K x^2$, where $K$ is the spring constant.
From this formula, it is clear that $U \propto x^2$.
When the stretch is increased from $x$ to $nx$, the new potential energy $U'$ becomes:
$U' = \frac{1}{2} K (nx)^2$
$U' = \frac{1}{2} K n^2 x^2$
$U' = n^2 (\frac{1}{2} K x^2)$
Since $U = \frac{1}{2} K x^2$, we have $U' = n^2 u$.

(D) The potential energy stored in a spring is given by $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the displacement.
This implies $U \propto x^2$.
Initially,the displacement is $x_1 = 2 \, cm$,so $U = C(2)^2 = 4C$,where $C$ is a constant.
When the spring is stretched by an additional $10 \, cm$,the new total displacement becomes $x_2 = 2 \, cm + 10 \, cm = 12 \, cm$.
The new energy stored is $U' = C(12)^2 = 144C$.
Taking the ratio of the new energy to the initial energy:
$\frac{U'}{U} = \frac{144C}{4C} = 36$.
Therefore,the new energy stored is $U' = 36U$.

(C) Let the original spring have length $\ell_0$ and force constant $k$. The spring constant is inversely proportional to the length,i.e.,$k \propto 1/\ell$.
When the spring is cut into two parts of lengths $\ell_1 = \ell_0/3$ and $\ell_2 = 2\ell_0/3$,their new spring constants are:
$k_1 = k \cdot (\ell_0 / \ell_1) = k \cdot (\ell_0 / (\ell_0/3)) = 3k$
$k_2 = k \cdot (\ell_0 / \ell_2) = k \cdot (\ell_0 / (2\ell_0/3)) = 1.5k = 3k/2$
Work done in stretching a spring by an amount $x$ is given by $W = \frac{1}{2} k x^2$.
Since both parts are stretched by the same amount $x$,the work done is directly proportional to the spring constant $(W \propto k)$.
For the shorter part (length $\ell_0/3$),$W_1 = \frac{1}{2} (3k) x^2 = 1.5 k x^2$.
For the longer part (length $2\ell_0/3$),$W_2 = \frac{1}{2} (1.5k) x^2 = 0.75 k x^2$.
Since $W_1 > W_2$,the work done is greater for the shorter part.

(A) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the spring force.
The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.
The final speed is $v_f = \frac{v}{2}$,so the final kinetic energy is $K_f = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{8}mv^2$.
The work done by the spring force when it is compressed by distance $x$ is $W = -\frac{1}{2}kx^2$.
Applying the theorem: $K_f - K_i = W$.
$\frac{1}{8}mv^2 - \frac{1}{2}mv^2 = -\frac{1}{2}kx^2$.
Multiplying by $8$: $mv^2 - 4mv^2 = -4kx^2$.
$-3mv^2 = -4kx^2$.
Solving for $k$: $k = \frac{3mv^2}{4x^2}$.

(A) According to the law of conservation of mechanical energy,the potential energy stored in the compressed spring is converted into the kinetic energy of the ball.
The potential energy of the spring is given by $U = \frac{1}{2} kx^2$.
The kinetic energy of the ball is given by $K = \frac{1}{2} mv^2$.
Given:
Mass $m = 15\, g = 15 \times 10^{-3}\, kg$
Force constant $k = 600\, N/m$
Compression $x = 3\, cm = 3 \times 10^{-2}\, m$
Equating the energies:
$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$
$kx^2 = mv^2$
$v^2 = \frac{k}{m} x^2$
$v = x \sqrt{\frac{k}{m}}$
Substituting the values:
$v = (3 \times 10^{-2}) \sqrt{\frac{600}{15 \times 10^{-3}}}$
$v = (3 \times 10^{-2}) \sqrt{\frac{600000}{15}}$
$v = (3 \times 10^{-2}) \sqrt{40000}$
$v = (3 \times 10^{-2}) \times 200$
$v = 6\, m/s$

(C) Let $m$ be the mass of ball $A$. When ball $A$ is released from rest,it moves downwards,stretching the spring. Let the spring stretch by a distance $x$ at the lowest point of the ball's motion.
By the law of conservation of energy,the loss in gravitational potential energy of ball $A$ is equal to the gain in elastic potential energy of the spring:
$mgx = \frac{1}{2}kx^2$
$mg = \frac{1}{2}kx$
$kx = 2mg$ ... $(1)$
For block $B$ of mass $M$ to leave contact with the ground,the upward spring force must be at least equal to the weight of block $B$:
$kx \ge Mg$ ... $(2)$
Substituting the value of $kx$ from equation $(1)$ into equation $(2)$:
$2mg \ge Mg$
$m \ge M/2$
Thus,the minimum mass of $A$ required is $M/2$.

(B) The work done by an external agent in stretching a spring is equal to the change in the elastic potential energy of the spring.
Given,spring constant $k = 1\, N/m$ and displacement $x = 2\, m$.
The work done $W$ is given by the formula:
$W = \int_{0}^{x} kx\, dx = \frac{1}{2} k x^2$
Substituting the values:
$W = \frac{1}{2} \times 1\, N/m \times (2\, m)^2$
$W = \frac{1}{2} \times 1 \times 4 = 2\, J$.
Therefore,the work done by Saroj is $2\, J$.

(D) The potential energy $U$ of a spring is given by the formula $U = \frac{1}{2} K x^2$,where $K$ is the spring constant and $x$ is the extension or compression.
This equation shows that $U \propto x^2$.
The graph of $U$ versus $x$ represents a parabola,not a straight line.
Therefore,the $Assertion$ is incorrect because the relationship is quadratic,not linear.
The $Reason$ is correct because it correctly states that the potential energy is proportional to the square of the extension or compression $(U \propto x^2)$.

(D) At maximum compression,the kinetic energy of the car is converted entirely into the potential energy of the spring.
First,convert the speed from $km/h$ to $m/s$:
$v = 18.0 \; km/h = 18.0 \times \frac{5}{18} \; m/s = 5 \; m/s$.
The kinetic energy $K$ of the moving car is:
$K = \frac{1}{2} m v^2 = \frac{1}{2} \times 1000 \; kg \times (5 \; m/s)^2 = 500 \times 25 = 1.25 \times 10^4 \; J$.
According to the law of conservation of mechanical energy,at maximum compression $x_m$,the potential energy $V$ of the spring equals the kinetic energy $K$ of the car:
$V = \frac{1}{2} k x_m^2 = K$.
Substituting the values:
$\frac{1}{2} \times (6.25 \times 10^3) \times x_m^2 = 1.25 \times 10^4$.
$x_m^2 = \frac{2 \times 1.25 \times 10^4}{6.25 \times 10^3} = \frac{2.5 \times 10^4}{6.25 \times 10^3} = 0.4 \times 10 = 4$.
$x_m = \sqrt{4} = 2 \; m$.

(N/A) Consider an elastic spring,obeying Hooke's law with negligible mass,whose one end is tied rigidly to a wall as shown in the figure. At the other end of the spring,a block is attached,resting on a smooth horizontal surface.
We shall,for the sake of simplicity,restrict the motion of the block to the $X$-direction.
In the normal position of the spring,the position of the block is taken as $x=0$,as shown in figure $(a)$.
When the block is pulled and the length of the spring increases by $x$,a restoring force $F_{S}$ is produced in the spring which tries to bring the spring back to its normal position. The restoring force (spring force) is also produced when the spring is compressed. This is shown in figures $(b)$ and $(c)$. The restoring force is directly proportional to the change in the length of the spring and is in the direction opposite to the change in length. This law of force for the spring is called Hooke's law: $F_{S} = -kx$,where $k$ is the spring constant or force constant of the spring.
Spring constant $k = \frac{|F_{S}|}{|x|}$,its unit is $N/m$.
The spring is said to be stiff if $k$ is large and soft if $k$ is small.
Suppose the block is pulled outwards as in figure $(b)$. If the extension is $x_{m}$,the work done by the spring force is:
$W_{S} = \int_{0}^{x_{m}} F_{S} dx = \int_{0}^{x_{m}} (-kx) dx$
$W_{S} = -k \left[ \frac{x^{2}}{2} \right]_{0}^{x_{m}} = -\frac{1}{2} k x_{m}^{2}$
The potential energy $V(x)$ of the spring is defined as the work done by an external force to stretch or compress it,which is the negative of the work done by the spring force:
$V(x) = -W_{S} = \frac{1}{2} k x^{2}$

(N/A) Consider a block of mass $m$ attached to a spring of spring constant $k$ on a frictionless horizontal surface. Let $x=0$ be the equilibrium position.
When the block is pulled to a displacement $x_m$ and released,it oscillates between $-x_m$ and $+x_m$. At any displacement $x$,the total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$:
$E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
At the extreme position $x = x_m$,the velocity $v = 0$,so the total energy is purely potential:
$E = \frac{1}{2}kx_m^2$
Since the system is conservative,the total energy remains constant at any position $x$:
$\frac{1}{2}kx_m^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
At the equilibrium position $x=0$,the potential energy is zero,and the kinetic energy is maximum,where $v = v_m$:
$\frac{1}{2}mv_m^2 = \frac{1}{2}kx_m^2$
Solving for $v_m$:
$v_m^2 = \frac{k}{m}x_m^2 \implies v_m = \sqrt{\frac{k}{m}}x_m$
This confirms that the kinetic energy is converted into potential energy and vice versa,while the total mechanical energy remains constant throughout the motion.

(N/A) The spring constant $(k)$ is a measure of the stiffness of a spring. It is defined as the force required to produce a unit extension or compression in the spring. According to Hooke's Law,$F = -kx$,where $F$ is the restoring force,$x$ is the displacement from the mean position,and $k$ is the spring constant. Its $SI$ unit is $N/m$.
The work done by a spring when it is displaced from position $x_1$ to $x_2$ is given by the formula: $W = -\frac{1}{2} k (x_2^2 - x_1^2)$.
From this expression,we can conclude that the work done by a spring depends on:
$1$. The spring constant $(k)$,which represents the stiffness of the spring.
$2$. The initial position $(x_1)$ of the spring.
$3$. The final position $(x_2)$ of the spring.

(C) The spring force is a conservative force.
By definition,the work done by a conservative force in a closed path or a cyclic process is always zero.
Since the spring force $F = -kx$ depends only on the displacement from the equilibrium position,the total work done over a complete cycle where the spring returns to its initial position is $\oint F \cdot dx = 0$.

(A) When an elastic spring is stretched or compressed,work is done against the restoring force of the spring.
According to the work-energy theorem,the work done by an external agent is stored in the spring in the form of elastic potential energy.
The potential energy of a spring is given by $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.
Since the potential energy depends on the square of the displacement $(x^2)$,any displacement (whether stretching or compressing) results in an increase in potential energy.

(A) When a spring is stretched or compressed by a displacement $x$ from its equilibrium position,work is done against the restoring force $F = -kx$. This work is stored in the spring as elastic potential energy. The potential energy $U$ is given by the formula $U = \frac{1}{2}kx^2$,where $k$ is the spring constant. Since $U$ is proportional to the square of the displacement $x^2$,any non-zero displacement (whether positive for stretching or negative for compression) results in an increase in the stored potential energy compared to the equilibrium state $(x=0)$.

(A) The work done by the stretching force is positive.
When a spring is stretched,the external stretching force is applied in the direction of the displacement of the spring's end.
Since the force $F$ and the displacement $dx$ are in the same direction,the work done $W = \int F \cdot dx$ is positive.

(C) Given mass $m = 50,000 \, kg$ and initial velocity $v = 36 \, km/h = 10 \, m/s$.
The initial kinetic energy $K$ is given by $K = \frac{1}{2} m v^2 = \frac{1}{2} \times 50,000 \times (10)^2 = 2.5 \times 10^6 \, J$.
Since $90 \%$ of the kinetic energy is lost due to friction,only $10 \%$ of the energy is stored in the spring of the shock absorber.
Energy stored in the spring $\Delta E = \frac{1}{2} k x^2$,where $x = 1.0 \, m$ is the compression.
According to the problem,$\Delta E = 10 \% \text{ of } K = 0.10 \times 2.5 \times 10^6 \, J = 2.5 \times 10^5 \, J$.
Equating the two expressions for energy: $\frac{1}{2} k (1.0)^2 = 2.5 \times 10^5 \, J$.
Solving for $k$: $k = 2 \times 2.5 \times 10^5 = 5.0 \times 10^5 \, N/m$.

(B) The potential energy $U$ stored in a spring stretched by a distance $y$ is given by $U = \frac{1}{2} k y^2$,where $k$ is the spring constant.
For the first case,$y_1 = 3 \, cm$,so $U_1 = \frac{1}{2} k (3)^2 = \frac{9}{2} k$.
For the second case,$y_2 = 6 \, cm$,so $U_2 = \frac{1}{2} k (6)^2 = \frac{36}{2} k$.
Taking the ratio of the two energies: $\frac{U_2}{U_1} = \frac{\frac{1}{2} k (6)^2}{\frac{1}{2} k (3)^2} = \frac{36}{9} = 4$.
Therefore,$U_2 = 4 U_1$.

(D) Let $x_{max}$ be the maximum extension of the spring.
By 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.
$mg x_{max} = \frac{1}{2} k x_{max}^2$
Solving for $x_{max}$ (assuming $x_{max} \neq 0$):
$x_{max} = \frac{2mg}{k}$
The maximum tension in the spring is given by Hooke's Law:
$T_{max} = k x_{max}$
Substituting the value of $x_{max}$:
$T_{max} = k \left( \frac{2mg}{k} \right) = 2mg$

(C) According to the law of conservation of mechanical energy,the kinetic energy of the ball is converted into the potential energy of the spring at the point of maximum compression.
Let the compression in the spring be $y$.
By the work-energy theorem or conservation of energy:
$\frac{1}{2} mv^2 = \frac{1}{2} ky^2$
Substituting the given values: $m = 4\, kg$,$v = 10\, ms^{-1}$,$k = 100\, Nm^{-1}$.
$\frac{1}{2} \times 4 \times (10)^2 = \frac{1}{2} \times 100 \times y^2$
$2 \times 100 = 50 \times y^2$
$200 = 50 \times y^2$
$y^2 = 4$
$y = 2\, m$
The initial length of the spring is $8\, m$. The compressed length $x$ is given by:
$x = \text{Initial length} - \text{Compression}$
$x = 8 - 2 = 6\, m$.
Thus,the value of $x$ is $6$.

(D) Let the mass of the ball be $m = 100 \, g = 0.1 \, kg$. The height from which it is dropped is $h = 10 \, cm = 0.1 \, m$. The compression in the spring is $x = \frac{h}{2} = 5 \, cm = 0.05 \, m$.
By the principle of conservation of mechanical energy,the loss in gravitational potential energy of the ball equals the gain in elastic potential energy of the spring.
The total vertical displacement of the ball is $h + x = h + \frac{h}{2} = \frac{3h}{2}$.
Therefore,$mg \left( h + \frac{h}{2} \right) = \frac{1}{2} kx^2$.
Substituting the values: $0.1 \times 10 \times \left( 0.1 + 0.05 \right) = \frac{1}{2} \times k \times (0.05)^2$.
$1 \times 0.15 = \frac{1}{2} \times k \times 0.0025$.
$0.15 = k \times 0.00125$.
$k = \frac{0.15}{0.00125} = \frac{150000}{1250} = 120 \, N \, m^{-1}$.

(B) Let the mass of each block be $m = 250\,g = 0.25\,kg$ and the spring constant be $k = 2\,N/m$.
At the moment of maximum elongation $x$,both blocks momentarily come to rest in the center-of-mass frame.
By the law of conservation of energy,the initial kinetic energy of the system is converted into the elastic potential energy of the spring at maximum elongation.
The total initial kinetic energy is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
The potential energy stored in the spring at maximum elongation $x$ is $U_f = \frac{1}{2}kx^2$.
Equating the two: $mv^2 = \frac{1}{2}kx^2$.
Solving for $x$: $x^2 = \frac{2mv^2}{k} \implies x = v \sqrt{\frac{2m}{k}}$.
Substituting the given values $m = 0.25\,kg$ and $k = 2\,N/m$:
$x = V \sqrt{\frac{2 \times 0.25}{2}} = V \sqrt{0.25} = 0.5V = \frac{V}{2}$.

(C) The work done in stretching a spring by a distance $x$ is given by the potential energy stored,$U = \frac{1}{2} K x^2$.
Initial work done to stretch the spring by distance $a$ is $W_1 = \frac{1}{2} K a^2$.
Total work done to stretch the spring by a total distance of $(a+b)$ is $W_2 = \frac{1}{2} K (a+b)^2$.
The work done in stretching the spring further by distance $b$ is the difference between the total work and the initial work:
$\Delta W = W_2 - W_1$
$\Delta W = \frac{1}{2} K (a+b)^2 - \frac{1}{2} K a^2$
$\Delta W = \frac{1}{2} K (a^2 + 2ab + b^2 - a^2)$
$\Delta W = \frac{1}{2} K (2ab + b^2)$
$\Delta W = \frac{1}{2} K b(2a + b)$.

(B) The elastic force is a conservative force. The work done by an external agent to stretch a spring is equal to the change in its potential energy.
The potential energy of a spring stretched by a length $x$ is given by $U = \frac{1}{2} kx^2$.
Initially,the string is stretched by length $x$. The initial potential energy is $U_i = \frac{1}{2} kx^2$.
After stretching it further by length $y$,the total extension becomes $(x+y)$. The final potential energy is $U_f = \frac{1}{2} k(x+y)^2$.
The work done by the external agent in the second stretching is the change in potential energy:
$W = U_f - U_i$
$W = \frac{1}{2} k(x+y)^2 - \frac{1}{2} kx^2$
$W = \frac{1}{2} k(x^2 + y^2 + 2xy) - \frac{1}{2} kx^2$
$W = \frac{1}{2} kx^2 + \frac{1}{2} ky^2 + kxy - \frac{1}{2} kx^2$
$W = \frac{1}{2} ky^2 + kxy$
$W = \frac{1}{2} ky(y + 2x)$
Thus,the work done is $\frac{1}{2} ky(2x+y)$.

(A) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the spring constant.
For the initial stretch $x_1 = 2\,cm$,the potential energy is $U = \frac{1}{2} k (2)^2 = 2k$.
For the final stretch $x_2 = 8\,cm$,the new potential energy $U'$ is $U' = \frac{1}{2} k (8)^2 = 32k$.
Dividing the two equations:
$\frac{U'}{U} = \frac{\frac{1}{2} k (8)^2}{\frac{1}{2} k (2)^2} = \left(\frac{8}{2}\right)^2 = (4)^2 = 16$.
Therefore,$U' = 16\,U$.

(C) When the block $B$ is displaced by a distance $x$ towards wall $1$,spring $S1$ is compressed by $x$,while spring $S2$ remains unstretched because support $M2$ is not fixed to the wall and moves with the block. The potential energy stored in the system is $U_i = \frac{1}{2} k x^2$.
When the block is released,it moves towards the equilibrium position and then towards wall $2$. When it moves towards wall $2$ by a distance $y$,spring $S2$ is compressed by $y$,while spring $S1$ remains unstretched. The potential energy stored in the system is $U_f = \frac{1}{2} (4k) y^2$.
Since there is no friction and the supports have negligible mass,the total mechanical energy is conserved.
Equating the initial and final potential energies: $\frac{1}{2} k x^2 = \frac{1}{2} (4k) y^2$.
Simplifying this,we get $x^2 = 4y^2$,which implies $y^2 = \frac{x^2}{4}$.
Taking the square root,we find $y = \frac{x}{2}$.
Therefore,the ratio $\frac{y}{x} = \frac{1}{2}$.

(A) The work done $W$ in stretching a spring by an extension $x$ is given by the formula $W = \frac{1}{2} k x^2$.
Since the extension $x$ is the same for both springs,the work done is directly proportional to the spring constant $k$ $(W \propto k)$.
Given that $k_{A} > k_{B}$,it follows that $W_{A} > W_{B}$.
Therefore,more work is required to stretch spring $A$.

(B) The potential energy $U$ stored in a spring is given by the formula $U = \frac{1}{2} K x^2$,where $K$ is the spring constant and $x$ is the extension.
For the first case,$x_1 = 1 \ cm$,so $U = \frac{1}{2} K (1)^2 = \frac{1}{2} K$.
For the second case,$x_2 = 4 \ cm$,so the new potential energy $U'$ is $U' = \frac{1}{2} K (4)^2 = 16 \times (\frac{1}{2} K)$.
Substituting $U$ into the equation for $U'$,we get $U' = 16 U$.