A spring having a force constant of 200 N/m rests on a frictionless horizontal surface. One end is in contact with a stationary wall, and a 4-kg block is pushed against the other end, compressing the spring 0.1 m. The block is then released with no initial velocity. What is the block's speed when the spring returns to its uncompressed length?

1 answer

To solve for the block's speed when the spring returns to its uncompressed length, we can use the concept of energy conservation.

  1. Calculate the potential energy stored in the compressed spring: The potential energy \( U \) stored in a compressed or stretched spring is given by the formula:

    \[ U = \frac{1}{2} k x^2 \]

    where:

    • \( k = 200 , \text{N/m} \) (the spring constant),
    • \( x = 0.1 , \text{m} \) (the compression of the spring).

    Plugging in the values:

    \[ U = \frac{1}{2} (200 , \text{N/m}) (0.1 , \text{m})^2 = \frac{1}{2} (200) (0.01) = 1 , \text{J} \]

  2. Use energy conservation to find the kinetic energy of the block: When the spring pushes the block back to its uncompressed position, all the potential energy stored in the spring is converted into kinetic energy \( K \) of the block (since there are no other forces doing work, such as friction).

    The kinetic energy \( K \) of the block is given by the formula:

    \[ K = \frac{1}{2} mv^2 \]

    where \( m = 4 , \text{kg} \) is the mass of the block, and \( v \) is its speed when the spring is uncompressed. According to the conservation of energy:

    \[ U = K \]

    Thus,

    \[ 1 , \text{J} = \frac{1}{2} (4 , \text{kg}) v^2 \]

  3. Solve for the speed \( v \): Rearranging the equation to solve for \( v^2 \):

    \[ 1 = 2 v^2 \quad \Rightarrow \quad v^2 = \frac{1}{2} \]

    Taking the square root of both sides:

    \[ v = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 , \text{m/s} \]

Thus, the speed of the block when the spring returns to its uncompressed length is approximately 0.707 m/s.