wPE O14

Solution 14

Solution to Question 14

m l (relaxed) P F_hand P (final)

1. Analyzing the Extension

Let the total relaxed length of the cord be $l$. The mark P is at a height $0.8l$ from the lower end (block). This means the segment of the cord between the block and P has a relaxed length $l_{BP, 0} = 0.8l$.

The upper end is raised until mark P reaches the position where the top of the relaxed cord initially was (height $l$ from the ground). At the moment the block leaves the ground, the block is still effectively at height 0.

  • Current length of segment BP: $l_{BP} = l$.
  • Relaxed length of segment BP: $l_{BP, 0} = 0.8l$.
  • Extension of segment BP: $\Delta l_{BP} = l – 0.8l = 0.2l$.

Since the cord is uniform, the strain is constant throughout:

$$ \text{Strain} = \frac{\Delta l_{BP}}{l_{BP, 0}} = \frac{0.2l}{0.8l} = \frac{1}{4} $$

Therefore, the total extension of the whole cord ($\Delta l_{total}$) is:

$$ \Delta l_{total} = \text{Strain} \times l = \frac{1}{4} l $$

2. Finding the Stiffness Constant (k)

The block leaves the ground when the tension in the cord ($T$) equals the weight of the block ($mg$). Using Hooke’s Law $T = k \Delta l_{total}$:

$$ k \left(\frac{l}{4}\right) = mg \implies k = \frac{4mg}{l} $$

3. Calculating Work Done

The work done by the hand is equal to the energy stored in the elastic cord plus any change in gravitational potential energy of the block. Since the block is just about to leave the ground, its potential energy change is zero. The kinetic energy change is also zero (slowly raised).

$$ W = U_{elastic} = \frac{1}{2} k (\Delta l_{total})^2 $$ $$ W = \frac{1}{2} \left(\frac{4mg}{l}\right) \left(\frac{l}{4}\right)^2 $$ $$ W = \frac{1}{2} \cdot \frac{4mg}{l} \cdot \frac{l^2}{16} = \frac{mgl}{8} $$

Correct Option: (b)

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