Work Energy and Power

Practice Questions:

Question 131: moderate

Two springs $A$ and $B$ having spring constant $K_A$ and $K_B$ ($K_A = 2K_B$) are stretched by applying force of equal magnitude. If energy stored in spring $A$ is $E$ then energy stored in $B$ will be:

(2001)

1. $2E$

2. $E/4$

3. $E/2$

4. $4E$

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Concept: Energy stored in a spring under constant force. Formula: $PE = \frac{F^2}{2K}$. When the same force $F$ is applied, potential energy is inversely proportional to the spring constant ($PE \propto 1/K$). Given $K_A = 2K_B$. The ratio $\frac{PE_B}{PE_A} = \frac{K_A}{K_B}$. Substituting $K_A = 2K_B$, we get $\frac{PE_B}{E} = \frac{2K_B}{K_B} = 2$. Therefore, $PE_B = 2E$.

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Question 132: moderate

A block of mass $10\text{ kg}$ moving in $x$ direction with a constant speed of $10\text{ m s}^{-1}$, is subjected to a retarding force $F = -0.1x\text{ J/m}$ during its travel from $x = 20\text{ m}$ to $30\text{ m}$. Its final K.E. will be: (2015)

1. $450\text{ J}$

2. $275\text{ J}$

3. $250\text{ J}$

4. $475\text{ J}$

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Initial kinetic energy is $K_i = \frac{1}{2}mv^2 = 500\text{ J}$. Work done by the retarding force is $W = \int_{20}^{30} (-0.1x)dx = -25\text{ J}$. Using work-energy theorem, final K.E. $K_f = K_i + W = 500 - 25 = 475\text{ J}$.

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