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

(C) The work done $W$ in stretching a spring by a force $F$ is given by the formula $W = \frac{F^2}{2k}$.
Since both springs are stretched by the same force $F$,the work done is inversely proportional to the force constant,i.e.,$W \propto \frac{1}{k}$.
Given that $k_1 > k_2$,it follows that $\frac{1}{k_1} < \frac{1}{k_2}$.
Therefore,$W_1 < W_2$.
This implies that more work is done in the case of the second spring.

(A) The potential energy stored in a spring is given by $U = \frac{1}{2}kx^2$.
Given: Force constant $k = 10\, N/m$,initial stretch $x_1 = 0.20\, m$,and final stretch $x_2 = 0.25\, m$.
The increase in potential energy is $\Delta U = U_2 - U_1 = \frac{1}{2}k(x_2^2 - x_1^2)$.
Substituting the values: $\Delta U = \frac{1}{2} \times 10 \times [(0.25)^2 - (0.20)^2]$.
$\Delta U = 5 \times (0.0625 - 0.0400)$.
$\Delta U = 5 \times 0.0225 = 0.1125\, J$.
Rounding to the nearest provided option,the increase in potential energy is about $0.1\, J$.

(A) The potential energy $U$ of a spring stretched by a distance $x$ is given by $U = \frac{1}{2} k x^2$.
Given that for $x = S$,$U_1 = \frac{1}{2} k S^2 = 10 \, J$.
We need to find the work done to stretch it by an additional distance $S$,meaning the final extension is $x_2 = S + S = 2S$.
The work done is equal to the change in potential energy: $W = U_2 - U_1$.
$U_2 = \frac{1}{2} k (2S)^2 = \frac{1}{2} k (4S^2) = 4 \times (\frac{1}{2} k S^2) = 4 \times 10 = 40 \, J$.
Therefore,$W = 40 \, J - 10 \, J = 30 \, J$.

(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.
Given $F = 10 \, N$ and $x = 1 \, mm = 1 \times 10^{-3} \, m$.
So,$k = \frac{10}{1 \times 10^{-3}} = 10^4 \, N/m$.
The work done $W$ in stretching the spring by a displacement $x = 40 \, mm = 40 \times 10^{-3} \, m$ is given by $W = \frac{1}{2} k x^2$.
Substituting the values: $W = \frac{1}{2} \times 10^4 \times (40 \times 10^{-3})^2$.
$W = \frac{1}{2} \times 10^4 \times (1600 \times 10^{-6}) = \frac{1}{2} \times 10^4 \times 1.6 \times 10^{-3} = 0.5 \times 16 = 8 \, J$.

(B) According to the law of conservation of mechanical energy,the kinetic energy of the body is converted into the potential energy of the spring at the point of maximum compression.
$K.E. = P.E._{spring}$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$
Given: $m = 0.1 \ kg$,$v = 10 \ m/s$,$k = 1000 \ N/m$.
Substituting the values:
$\frac{1}{2} \times 0.1 \times (10)^2 = \frac{1}{2} \times 1000 \times x^2$
$0.1 \times 100 = 1000 \times x^2$
$10 = 1000 \times x^2$
$x^2 = \frac{10}{1000} = 0.01$
$x = \sqrt{0.01} = 0.1 \ m$.

(B) 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 $k = 800\, N/m$.
Initial extension $x_1 = 5\, cm = 0.05\, m$.
Final extension $x_2 = 15\, cm = 0.15\, m$.
Substituting the values into the formula:
$W = \frac{1}{2} \times 800 \times ((0.15)^2 - (0.05)^2)$
$W = 400 \times (0.0225 - 0.0025)$
$W = 400 \times 0.0200$
$W = 8\, J$.

(C) The potential energy stored in a spring is given by $U = \frac{1}{2} k x^2$.
Initially,for $x_1 = 2 \, cm$,the energy $U_1 = 100 \, J$.
So,$100 = \frac{1}{2} k (2)^2 \implies 100 = 2k \implies k = 50 \, J/cm^2$.
When the spring is stretched further by $2 \, cm$,the new extension is $x_2 = 2 \, cm + 2 \, cm = 4 \, cm$.
The new potential energy is $U_2 = \frac{1}{2} k x_2^2 = \frac{1}{2} (50) (4)^2 = 25 \times 16 = 400 \, J$.
The increase in stored energy is $\Delta U = U_2 - U_1 = 400 \, J - 100 \, J = 300 \, J$.

(D) The potential energy $U$ of a spring stretched by a distance $x$ is given by the formula $U = \frac{1}{2}kx^2$,where $k$ is the spring constant.
Given,for $x_1 = 2 \,mm$,$U_1 = 4 \,J$.
When the spring is stretched by $x_2 = 10 \,mm$,the new potential energy $U_2$ is proportional to the square of the displacement $(U \propto x^2)$.
Therefore,$\frac{U_2}{U_1} = \left(\frac{x_2}{x_1}\right)^2$.
Substituting the values: $\frac{U_2}{4} = \left(\frac{10}{2}\right)^2 = (5)^2 = 25$.
Thus,$U_2 = 4 \times 25 = 100 \,J$.

(C) The work done 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:
Spring constant $k = 5 \times 10^3 \, N/m$.
Initial extension $x_1 = 5 \, cm = 0.05 \, m$.
Final extension $x_2 = 5 \, cm + 5 \, cm = 10 \, cm = 0.10 \, m$.
Substituting the values into the formula:
$W = \frac{1}{2} \times (5 \times 10^3) \times ((0.10)^2 - (0.05)^2)$
$W = \frac{1}{2} \times 5000 \times (0.01 - 0.0025)$
$W = 2500 \times 0.0075$
$W = 18.75 \, J$.

(A) According to the law of conservation of energy,the kinetic energy of the mass is converted into the elastic potential energy of the spring at the point of maximum compression.
Let $m = 0.5\,kg$ be the mass,$v = 1.5\,m/s$ be the velocity,and $k = 50\,N/m$ be the spring constant.
The kinetic energy of the mass is $K.E. = \frac{1}{2}mv^2$.
The potential energy stored in the spring at maximum compression $x$ is $P.E. = \frac{1}{2}kx^2$.
Equating the two energies:
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$
Solving for $x$:
$x^2 = \frac{mv^2}{k}$
$x = \sqrt{\frac{mv^2}{k}} = v\sqrt{\frac{m}{k}}$
Substituting the given values:
$x = 1.5 \times \sqrt{\frac{0.5}{50}}$
$x = 1.5 \times \sqrt{0.01}$
$x = 1.5 \times 0.1 = 0.15\,m$.
Thus,the maximum compression is $0.15\,m$.

(C) The potential energy $U$ of a spring stretched by a distance $x$ is given by the formula: $U = \frac{1}{2} k x^2$.
Since the spring constant $k$ is constant,the potential energy is directly proportional to the square of the displacement: $U \propto x^2$.
Initially,for $x_1 = 1 \, cm$,the potential energy is $U_1 = U$.
When the spring is stretched by $x_2 = 4 \, cm$,the new potential energy $U_2$ is:
$U_2 = \frac{1}{2} k (x_2)^2 = \frac{1}{2} k (4 \, cm)^2 = 16 \times (\frac{1}{2} k (1 \, cm)^2) = 16 U$.
Therefore,the potential energy becomes $16U$.

(B) The restoring force exerted by the spring is given by $F = -kx$,where $k$ is the spring constant and $x$ is the displacement.
To calculate the work done by an external agent to extend the spring,we apply an external force $F_{ext} = -F = kx$.
The work done $W$ is the integral of the force over the displacement:
$W = \int_{0}^{x_1} F_{ext} dx = \int_{0}^{x_1} kx dx$
$W = k \left[ \frac{x^2}{2} \right]_{0}^{x_1}$
$W = k \left( \frac{x_1^2}{2} - 0 \right)$
$W = \frac{1}{2} k x_1^2$
Thus,the work done in extending the spring is $\frac{1}{2} k x_1^2$.

(D) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by the formula $U = \frac{1}{2}kx^2$,where $k$ is the spring constant.
From this relation,it is clear that $U \propto x^2$.
Given that for $x_1 = 0.02\, m$,the potential energy is $U_1 = U$.
For $x_2 = 0.1\, m$,let the potential energy be $U_2$.
Using the proportionality,we have $\frac{U_2}{U_1} = \left( \frac{x_2}{x_1} \right)^2$.
Substituting the values: $\frac{U_2}{U} = \left( \frac{0.1}{0.02} \right)^2 = (5)^2 = 25$.
Therefore,$U_2 = 25U$.

(B) The potential energy $(PE)$ stored in a spring compressed or stretched by a distance $x$ is given by the formula:
$PE = \frac{1}{2} k x^2$
where $k$ is the spring constant.
Given that the spring is compressed by a distance $a$,we substitute $x = a$ into the formula:
$PE = \frac{1}{2} k a^2$
Since $\frac{1}{2}$ and $k$ are constants,the potential energy is directly proportional to the square of the displacement:
$PE \propto a^2$
Therefore,the correct option is $B$.

(B) According to the law of conservation of energy,the total mechanical energy of the system remains constant.
Let the initial position of the block be the reference level for gravitational potential energy $(U_g = 0)$.
When the block is at height $h$ above the spring,its total energy is $E_i = 0$ (since it is at rest and at the reference level).
When the spring is compressed by a maximum distance $x$,the block comes to rest momentarily at the lowest point.
The total vertical displacement of the block from its initial position is $(h + x)$.
The change in gravitational potential energy is $\Delta U_g = -mg(h + x)$.
The elastic potential energy stored in the spring is $U_s = \frac{1}{2}kx^2$.
Since the total energy is conserved,the loss in gravitational potential energy is equal to the gain in elastic potential energy:
$mg(h + x) = \frac{1}{2}kx^2$.

(C) When a block of mass $M$ moving with velocity $v$ collides with a spring,its kinetic energy is converted into the elastic potential energy of the spring at the point of maximum compression $L$.
By the law of conservation of energy:
$\frac{1}{2} M v^2 = \frac{1}{2} K L^2$
Solving for $v$:
$v^2 = \frac{K}{M} L^2 \implies v = L \sqrt{\frac{K}{M}}$
The momentum $P$ of the block is given by $P = Mv$.
Substituting the value of $v$:
$P = M \left( L \sqrt{\frac{K}{M}} \right) = \sqrt{M^2 \cdot \frac{K}{M}} L = \sqrt{MK} L$
Thus,the maximum momentum of the block is $\sqrt{MK} L$.

(D) The potential energy $U$ stored in a spring stretched by a distance $l$ is given by the formula $U = \frac{1}{2} k l^2$,where $k$ is the spring constant.
This implies that $U \propto l^2$.
Given that the initial potential energy $U_1 = V$ for an extension $l_1 = 2 \, cm$.
We need to find the potential energy $U_2$ for an extension $l_2 = 10 \, cm$.
Using the proportionality $U \propto l^2$,we have:
$\frac{U_2}{U_1} = \left( \frac{l_2}{l_1} \right)^2$
$\frac{U_2}{V} = \left( \frac{10}{2} \right)^2 = (5)^2 = 25$
Therefore,$U_2 = 25V$.

(B) The potential energy $U$ stored in a stretched spring is given by the formula $U = \frac{1}{2} k x^2$.
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 $B$.

(D) The potential energy stored in a spring with force constant $K$ at an extension $x$ is given by $U = \frac{1}{2}Kx^2$.
At extension $l_1$,the stored potential energy is $U_1 = \frac{1}{2}Kl_1^2$.
At extension $l_2$,the stored potential energy is $U_2 = \frac{1}{2}Kl_2^2$.
The work done $W$ in increasing the extension from $l_1$ to $l_2$ is equal to the change in potential energy:
$W = U_2 - U_1$
$W = \frac{1}{2}Kl_2^2 - \frac{1}{2}Kl_1^2$
$W = \frac{1}{2}K(l_2^2 - l_1^2)$.

(B) Given: Mass $m = 4\, kg$,extension $x_1 = 2\, cm = 0.02\, m$,final extension $x_2 = 5\, cm = 0.05\, m$,$g = 9.8\, m/s^2$.
First,find the spring constant $K$ using Hooke's law: $F = Kx_1 \implies mg = Kx_1$.
$K = \frac{mg}{x_1} = \frac{4 \times 9.8}{0.02} = \frac{39.2}{0.02} = 1960\, N/m$.
The work done by an external agent to stretch the spring by $x_2$ is given by $W = \frac{1}{2} K x_2^2$.
$W = \frac{1}{2} \times 1960 \times (0.05)^2$.
$W = 980 \times 0.0025 = 2.45\, J$.

(C) The elastic potential energy $U$ stored in a spring is given by $U = \frac{F^2}{2K}$,where $F$ is the stretching force and $K$ is the force constant.
Given that the elastic energies are equal,we have $U_1 = U_2$.
Therefore,$\frac{F_1^2}{2K_1} = \frac{F_2^2}{2K_2}$.
Rearranging the terms,we get $\frac{F_1^2}{F_2^2} = \frac{K_1}{K_2}$.
Taking the square root on both sides,we find $\frac{F_1}{F_2} = \sqrt{\frac{K_1}{K_2}}$.
Thus,the ratio $F_1:F_2$ is $\sqrt{K_1}:\sqrt{K_2}$.

(A) The potential energy $U$ stored in a spring stretched by a force $F$ is given by the formula $U = \frac{F^2}{2K}$,where $K$ is the spring constant.
Since the force $F$ is the same for both springs,we have $U \propto \frac{1}{K}$.
Therefore,the ratio of potential energy stored in the two springs is $\frac{U_1}{U_2} = \frac{K_2}{K_1}$.
Substituting the given values $K_1 = 1500 \, N/m$ and $K_2 = 3000 \, N/m$,we get $\frac{U_1}{U_2} = \frac{3000}{1500} = 2$.
Thus,the ratio is $2:1$.

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

(C) The potential energy $(P.E.)$ stored in a stretched spring is given by the formula $P.E. = \frac{1}{2}kh^2$,where $k$ is the spring constant and $h$ is the extension.
When the body is in equilibrium,the gravitational force acting downwards is balanced by the spring force acting upwards.
Thus,$mg = kh$,which implies $k = \frac{mg}{h}$.
Substituting the value of $k$ into the potential energy formula:
$P.E. = \frac{1}{2} \left( \frac{mg}{h} \right) h^2$.
$P.E. = \frac{1}{2} mgh$.

(C) The potential energy stored in a spring is given by $U = \frac{1}{2}kx^2$.
Initially,for $x_1 = 2 \ cm$,the energy $U_1 = 100 \ J = \frac{1}{2}k(2)^2$.
When the spring is stretched by another $2 \ cm$,the total extension becomes $x_2 = 2 \ cm + 2 \ cm = 4 \ cm$.
The total energy stored at $x_2$ is $U_2 = \frac{1}{2}k(4)^2 = \frac{1}{2}k(16)$.
Since $U_1 = \frac{1}{2}k(4) = 100 \ J$,we have $\frac{1}{2}k = 25 \ J/cm^2$.
Therefore,$U_2 = 25 \times 16 = 400 \ J$.
The additional energy stored is $\Delta U = U_2 - U_1 = 400 \ J - 100 \ J = 300 \ J$.

(D) The potential energy stored in a spring stretched by a length $x$ is given by $U_1 = \frac{1}{2}Kx^2$.
When the spring is further stretched by an additional length $y$,the total extension becomes $(x + y)$.
The total potential energy stored in the spring is $U_2 = \frac{1}{2}K(x + y)^2$.
The work done in the second stretching is the change in potential energy:
$W = U_2 - U_1$
$W = \frac{1}{2}K(x + y)^2 - \frac{1}{2}Kx^2$
$W = \frac{1}{2}K(x^2 + 2xy + y^2) - \frac{1}{2}Kx^2$
$W = \frac{1}{2}K(x^2 + 2xy + y^2 - x^2)$
$W = \frac{1}{2}K(2xy + y^2)$
$W = \frac{1}{2}Ky(2x + y)$.

(B) The work done $W$ in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$.
The work done in stretching the spring from an initial extension $x_1$ to a final extension $x_2$ is given by $W = \frac{1}{2} k (x_2^2 - x_1^2)$.
For the first case ($10 \ cm$ to $20 \ cm$): $W_1 = \frac{1}{2} k (20^2 - 10^2) = \frac{1}{2} k (400 - 100) = 150 k$.
For the second case ($20 \ cm$ to $30 \ cm$): $W_2 = \frac{1}{2} k (30^2 - 20^2) = \frac{1}{2} k (900 - 400) = 250 k$.
Comparing the two,$W_1 < W_2$. Therefore,the work done in stretching the spring from $10 \ cm$ to $20 \ cm$ is less than the work done in stretching it from $20 \ cm$ to $30 \ cm$.

(D) 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 displacement from the equilibrium position.
Since $x$ is squared in the formula,the potential energy $U$ is always positive for any non-zero displacement $(x \neq 0)$.
Whether the spring is stretched $(x > 0)$ or compressed $(x < 0)$,the value of $x^2$ increases as the magnitude of displacement increases.
Therefore,the potential energy of a spring increases whenever it is either stretched or compressed from its natural length.

(D) The work done $W$ on a spring with force constant $k$ when stretched by a force $F$ is given by $W = \frac{F^2}{2k}$.
Since the force $F$ is the same for both springs,the work done is inversely proportional to the force constant $k$ $(W \propto \frac{1}{k})$.
Given $k_1 < k_2$,it follows that $\frac{1}{k_1} > \frac{1}{k_2}$.
Therefore,$W_1 = \frac{F^2}{2k_1} > W_2 = \frac{F^2}{2k_2}$.
Thus,Statement-$1$ is true because the work done on $S_1$ is indeed greater than the work done on $S_2$.
Statement-$2$ $(k_1 < k_2)$ is also true and it is the direct reason why $W_1 > W_2$.
Hence,Statement-$2$ is the correct explanation for Statement-$1$.

(A) According to the law of conservation of energy,the kinetic energy of the body is converted into the elastic potential energy of the spring at maximum compression.
Let $m = 0.5 \ kg$,$v = 1.5 \ m/s$,$k = 50 \ N/m$,and $x$ be the maximum compression.
Kinetic Energy = Elastic Potential Energy
$\frac{1}{2} mv^2 = \frac{1}{2} kx^2$
$mv^2 = kx^2$
$0.5 \times (1.5)^2 = 50 \times x^2$
$0.5 \times 2.25 = 50 \times x^2$
$1.125 = 50 \times x^2$
$x^2 = \frac{1.125}{50} = 0.0225$
$x = \sqrt{0.0225} = 0.15 \ m$
Therefore,the maximum compression is $0.15 \ m$.

(C) The ball falls from a height $h$ and compresses the spring by a distance $x$.
According to the law of conservation of energy,the loss in gravitational potential energy of the ball is equal to the gain in elastic potential energy of the spring.
The total vertical displacement of the ball is $(h + x)$.
Therefore,the loss in potential energy is $mg(h + x)$.
The gain in elastic potential energy of the spring is $\frac{1}{2}kx^2$.
Equating the two: $mg(h + x) = \frac{1}{2}kx^2$.
Solving for the spring constant $k$: $k = \frac{2mg(h + x)}{x^2}$.

(B) Let $x$ be the maximum extension in the spring. At the point of maximum extension,the velocity of the block is zero.
Applying the principle of conservation of mechanical 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 $x \neq 0$): $x = \frac{2Mg}{k}$

(C) The potential energy $U$ stored in a spring is given by $U = \frac{1}{2} k x^2$.
Since the force $F$ applied is the same,we use the relation $F = kx$,which implies $x = \frac{F}{k}$.
Substituting this into the energy formula: $U = \frac{1}{2} k (\frac{F}{k})^2 = \frac{F^2}{2k}$.
Since $F$ is constant,$U \propto \frac{1}{k}$.
Therefore,the ratio of 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 $2 : 1$.

(D) Given: Natural length $L_0 = 10 \ cm$,extension $x_1 = 2 \ cm = 0.02 \ m$,mass $m = 20 \ g = 0.02 \ kg$,$g = 10 \ m/s^2$.
At equilibrium,the spring force equals the gravitational force: $k x_1 = mg$.
$k(0.02) = 0.02 \times 10 \implies k = 10 \ N/m$.
The total length becomes $14 \ cm$,so the new extension is $x_2 = 14 \ cm - 10 \ cm = 4 \ cm = 0.04 \ m$.
The elastic potential energy stored in the spring is given by $U = \frac{1}{2} k x_2^2$.
$U = \frac{1}{2} \times 10 \times (0.04)^2 = 5 \times 0.0016 = 0.008 \ J = 8 \times 10^{-3} \ J$.

(B) According to the law of conservation of energy,the potential energy lost by the block as it descends a vertical height $h$ is equal to the elastic potential energy gained by the spring at maximum compression $x$.
The loss in gravitational potential energy of the block is $\Delta U_g = mgh$.
The gain in elastic potential energy of the spring is $\Delta U_s = \frac{1}{2} Kx^2$.
Equating the two:
$mgh = \frac{1}{2} Kx^2$
Solving for $x$:
$x^2 = \frac{2mgh}{K}$
$x = \sqrt{\frac{2mgh}{K}}$

(A) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the elastic potential energy of the spring at the point of maximum compression.
Let $v$ be the velocity of the block just before it hits the spring.
$\frac{1}{2} M v^2 = \frac{1}{2} k L^2$
Solving for $v$:
$v^2 = \frac{k L^2}{M} \implies v = L \sqrt{\frac{k}{M}}$
The momentum $p$ of the block is given by $p = Mv$.
Substituting the value of $v$:
$p = M \left( L \sqrt{\frac{k}{M}} \right) = L \sqrt{M^2 \cdot \frac{k}{M}} = L \sqrt{Mk} = \sqrt{Mk} \, L$
Thus,the maximum momentum of the block is $\sqrt{Mk} \, L$.

(A) The potential energy stored in a spring stretched by distance $s$ is given by $U_1 = \frac{1}{2}ks^2 = 10 \ J$.
When the spring is stretched by an additional distance $s$,the total extension becomes $x = s + s = 2s$.
The potential energy at this new extension is $U_2 = \frac{1}{2}k(2s)^2 = \frac{1}{2}k(4s^2) = 4 \times (\frac{1}{2}ks^2)$.
Substituting $U_1 = 10 \ J$,we get $U_2 = 4 \times 10 \ J = 40 \ J$.
The work done to stretch the spring from $s$ to $2s$ is the change in potential energy: $W = U_2 - U_1 = 40 \ J - 10 \ J = 30 \ J$.

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

(D) The initial potential energy of the spring when stretched by $x$ is $U_1 = \frac{1}{2}kx^2$.
The final potential energy of the spring when stretched by a total distance $(x + y)$ is $U_2 = \frac{1}{2}k(x + y)^2$.
The work done during the second stretching is equal to the change in potential energy:
$W = U_2 - U_1$
$W = \frac{1}{2}k(x + y)^2 - \frac{1}{2}kx^2$
$W = \frac{1}{2}k(x^2 + 2xy + y^2 - x^2)$
$W = \frac{1}{2}k(2xy + y^2)$
$W = \frac{1}{2}ky(2x + y)$.

(C) According to the law of conservation of energy,the kinetic energy lost by the block is equal to the potential energy gained by the spring.
Let $d$ be the maximum compression in the spring.
The kinetic energy of the block is $K = \frac{1}{2}mv^2$,where $m = \frac{W}{g}$ is the mass of the block.
The potential energy stored in the spring at maximum compression is $U = \frac{1}{2}kd^2$.
Equating the two: $\frac{1}{2}mv^2 = \frac{1}{2}kd^2$.
Substituting $m = \frac{W}{g}$:
$\frac{1}{2} \left( \frac{W}{g} \right) v^2 = \frac{1}{2}kd^2$.
Solving for $d$:
$d^2 = \frac{Wv^2}{kg}$.
Therefore,$d = v\sqrt{\frac{W}{kg}}$.

(B) The force constant of the spring is $k = 100 \ N/m$.
The displacement of the spring is $x = 5 \ cm = 0.05 \ m$.
The work done in stretching a spring is given by the formula $W = \frac{1}{2} k x^2$.
Substituting the values into the formula:
$W = \frac{1}{2} \times 100 \times (0.05)^2$
$W = 50 \times 0.0025$
$W = 0.125 \ J$.

(B) The work done in stretching a spring is stored as potential energy in the spring,given by $U = \frac{1}{2} kx^2$.
The work done $W$ in stretching the spring from an initial extension $x_1$ to a final extension $x_2$ is given by the change in potential energy:
$W = \frac{1}{2} k x_2^2 - \frac{1}{2} k x_1^2 = \frac{1}{2} k (x_2^2 - x_1^2)$.
Given: $k = 800 \ N/m$,$x_1 = 5 \ cm = 0.05 \ m$,$x_2 = 15 \ cm = 0.15 \ m$.
$W = \frac{1}{2} \times 800 \times ((0.15)^2 - (0.05)^2)$
$W = 400 \times (0.0225 - 0.0025)$
$W = 400 \times 0.02 = 8 \ J$.

(C) The potential energy of a spring stretched by distance $x$ is given by $U_1 = \frac{1}{2}kx^2 = 10 \ J$.
When the spring is stretched by an additional distance $x$,the total extension becomes $x + x = 2x$.
The new potential energy is $U_2 = \frac{1}{2}k(2x)^2 = \frac{1}{2}k(4x^2) = 4 \times (\frac{1}{2}kx^2) = 4 \times 10 = 40 \ J$.
The work done to stretch the spring by the additional distance is the change in potential energy: $W = U_2 - U_1 = 40 \ J - 10 \ J = 30 \ J$.

(B) The work done to stretch a spring from an initial extension $x_1$ to a final extension $x_2$ is given by $W = \frac{1}{2} k (x_2^2 - x_1^2)$.
Here,$k = 5 \times 10^3 \ N/m$,$x_1 = 5 \ cm = 0.05 \ m$,and $x_2 = 5 \ cm + 5 \ cm = 10 \ cm = 0.10 \ m$.
Substituting the values:
$W = \frac{1}{2} \times (5 \times 10^3) \times [(0.10)^2 - (0.05)^2]$
$W = \frac{1}{2} \times 5000 \times [0.01 - 0.0025]$
$W = 2500 \times 0.0075$
$W = 18.75 \ J$.

(D) According to the law of conservation of mechanical energy,the potential energy at the initial height is converted into the elastic potential energy of the spring at maximum compression.
$mgh = \frac{1}{2} kx^2$
Given:
Mass $m = 40 \ g = 0.04 \ kg$
Height $h = 4.9 \ m$
Spring constant $k = 400 \ N/m$
Acceleration due to gravity $g = 9.8 \ m/s^2$
Substituting the values:
$0.04 \times 9.8 \times 4.9 = \frac{1}{2} \times 400 \times x^2$
$1.9208 = 200 \times x^2$
$x^2 = \frac{1.9208}{200} = 0.009604$
$x = \sqrt{0.009604} = 0.098 \ m$
Converting to centimeters: $x = 0.098 \times 100 = 9.8 \ cm$.

(A) By the law of conservation of mechanical energy,the potential energy lost by the particle is equal to the elastic potential energy gained by the spring.
$mg(h + x) = \frac{1}{2} kx^2$
For the first case: $m = 0.1 \; kg$,$h_1 = 0.24 \; m$,$x_1 = 0.01 \; m$.
$mg(0.24 + 0.01) = \frac{1}{2} k(0.01)^2$ --- $(1)$
For the second case: $h_2 = h$,$x_2 = 0.04 \; m$.
$mg(h + 0.04) = \frac{1}{2} k(0.04)^2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{h + 0.04}{0.24 + 0.01} = \frac{(0.04)^2}{(0.01)^2}$
$\frac{h + 0.04}{0.25} = \frac{0.0016}{0.0001} = 16$
$h + 0.04 = 16 \times 0.25 = 4$
$h = 4 - 0.04 = 3.96 \; m$

(A) According to the law of conservation of mechanical energy,the initial potential energy stored in the spring is converted into the kinetic energy of the block when the spring reaches its natural length.
Initial compression of the spring,$x = \frac{{\ell _0}}{2}$.
Initial potential energy,$U_i = \frac{1}{2}k x^2 = \frac{1}{2}k \left( \frac{{\ell _0}}{2} \right)^2 = \frac{1}{8}k \ell_0^2$.
Final kinetic energy of the block,$K_f = \frac{1}{2}mv^2$.
By conservation of energy,$U_i = K_f$.
$\frac{1}{8}k \ell_0^2 = \frac{1}{2}mv^2$.
$v^2 = \frac{k \ell_0^2}{4m}$.
$v = \frac{{\ell _0}}{2}\sqrt {\frac{k}{m}}$.

(C) According to the law of conservation of energy,the potential energy of the particle at height $h$ plus the compression $x$ is converted into the elastic potential energy of the spring:
$mg(h + x) = \frac{1}{2} K x^2$
Case $I$: $m = 0.1 \ kg$,$h_1 = 0.24 \ m$,$x_1 = 0.01 \ m$
$mg(0.24 + 0.01) = \frac{1}{2} K (0.01)^2$
$mg(0.25) = \frac{1}{2} K (0.0001) \quad ... (1)$
Case $II$: $m = 0.1 \ kg$,$h_2 = h$,$x_2 = 0.04 \ m$
$mg(h + 0.04) = \frac{1}{2} K (0.04)^2$
$mg(h + 0.04) = \frac{1}{2} K (0.0016) \quad ... (2)$
Dividing equation $(2)$ by $(1)$:
$\frac{h + 0.04}{0.25} = \frac{0.0016}{0.0001} = 16$
$h + 0.04 = 16 \times 0.25$
$h + 0.04 = 4$
$h = 4 - 0.04 = 3.96 \ m$

(A) Let the speed of the block of mass $m$ be $v$ and the speed of the block of mass $M$ be $V$.
Since the system is on a frictionless table,the total mechanical energy and linear momentum are conserved.
According to the law of conservation of mechanical energy:
Initial potential energy = Final kinetic energy
$\frac{1}{2} K x^2 = \frac{1}{2} m v^2 + \frac{1}{2} M V^2 \quad \dots(1)$
According to the law of conservation of linear momentum (initial momentum is zero):
$m v = M V \implies v = \frac{M V}{m} \quad \dots(2)$
Substituting $v$ from equation $(2)$ into equation $(1)$:
$K x^2 = m \left( \frac{M V}{m} \right)^2 + M V^2$
$K x^2 = \frac{M^2 V^2}{m} + M V^2 = M V^2 \left( \frac{M}{m} + 1 \right) = M V^2 \left( \frac{M+m}{m} \right)$
$V^2 = \frac{K x^2 m}{M(M+m)}$
$V = \sqrt{\frac{Km}{M(M+m)}} \cdot x$

(D) 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 displacement.
From this relation,we can see that $U \propto x^2$.
Given the initial displacement $x_1 = 2 \ cm$ and initial energy $U_1 = U$.
The new displacement is $x_2 = 10 \ cm$.
Using the ratio method: $\frac{U_2}{U_1} = \left( \frac{x_2}{x_1} \right)^2$.
Substituting the values: $\frac{U_2}{U} = \left( \frac{10 \ cm}{2 \ cm} \right)^2 = (5)^2 = 25$.
Therefore,the new potential energy is $U_2 = 25 U$.