Question

A block is kept on a smooth surface in given arrangement. Length and stiffness constant (k) of each spring is same.

The block (of mass m) is pulled to the right by 5 m and then released from rest. Maximum speed reached by block will be (assume k = 10 N/m and m = 5 kg)

• A. 5 m/s
• B. 5√2 m/s
• C. 10 m/s
• D. 10√2 m/s

Answer 1

Answer: (C)

10 m/s

Answer 2

The total potential energy stored in both springs is given by $U = kx^2$, where $k$ is the spring constant and $x$ is the displacement.

When the block is released, this potential energy is converted into kinetic energy, given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity.

Equating the potential and kinetic energies:

$$\frac{1}{2}mv^2 = kx^2$$

Solving for $v$:

$$v = \sqrt{\frac{2kx^2}{m}}$$

Given $k = 10 \, \text{N/m}$, $m = 5 \, \text{kg}$, and $x = 5 \, \text{m}$:

$$v = \sqrt{\frac{2 \times 10 \times 5^2}{5}} = \sqrt{\frac{500}{5}} = \sqrt{100} = 10 \, \text{m/s}$$

Thus, the maximum speed reached by the block is $10 \, \text{m/s}$.