Question

In the system shown in the figure, initially both springs are relaxed. Left block is slowly pulled down by 0.1 m and released. Maximum value of $K_1$ for which both blocks will have same magnitude of acceleration just after release is ______ N/m. Take $(g = 10 m/s^2)$

Answer 1

200 N/m

Answer 2

Solution:

Let upward be positive for both blocks. Initially the left block is displaced downward by 0.1 m so its displacement is $x_1=-0.1\,\text{m}$ while the right block is at $x_2=0$. (Both springs are relaxed at the natural length.) Each block is of mass 1 kg and is connected by an ideal string over a fixed pulley; hence their motions are related by

$$a_1 = -a_2.$$

Denote the common magnitude of acceleration as $a$ (with $a>0$ meaning the left block accelerates upward and the right block accelerates downward). The spring forces act upward (restoring) when a block is displaced downward. Writing Newton’s second law for each block:

Left block:
Its forces (all taken positive upward) are the spring force $K_1\,(0.1)$ (because a downward displacement of 0.1 m stretches the spring, so it pulls upward), plus the rope tension $T$ upward, while its weight $mg$ (with $g=10\,\text{m/s}^2$) acts downward. Hence,

$$T + K_1(0.1) - mg = m\,a_1.$$

Since $m=1,\;g=10$ and $a_1 = a$ (upward),

$$T + 0.1\,K_1 - 10 = a. \qquad (1)$$

Right block:
Its spring is not stretched (since $x_2=0$) so it provides no force. Taking upward as positive, weight acts downward and tension $T$ upward. Its acceleration is $a_2$, but by the constraint $a_2=-a$ (i.e. it accelerates downward),

$$T - 10 = m\,a_2 = -a. \qquad (2)$$

Equating (1) and (2):
Solve (2) for $T$:

$$T = 10 - a.$$

Substitute into (1):

$$(10 - a) + 0.1\,K_1 - 10 = a \quad\Longrightarrow\quad 0.1\,K_1 - a = a.$$

Thus,

$$0.1\,K_1 = 2a \quad\Longrightarrow\quad a = 0.05\,K_1.$$

Now, the tension from (2) is:

$$T = 10 - 0.05\,K_1.$$

For the string to remain taut (i.e. $T\ge 0$), we require:

$$10 - 0.05\,K_1 \ge 0 \quad\Longrightarrow\quad K_1 \le 200\,\text{N/m}.$$

Thus, the maximum value of $K_1$ for which both blocks have the same magnitude of acceleration immediately after release is $200\,\text{N/m}$.

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Core Explanation (Minimal):

1. Write equations of motion (using upward-positive) for the left block: $$T + 0.1\,K_1 - 10 = a.$$
2. For the right block (with no spring force): $$T - 10 = -a.$$
3. Solve to obtain $a = 0.05\,K_1$ and $T=10-0.05\,K_1$.
4. For the rope to be taut, $T\ge0 \Rightarrow 10-0.05\,K_1\ge0$, hence $K_1\le200\,\text{N/m}$.