Question

A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0. 75 m. Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it? 3. 5 m/s 4. 1 m/s 13 m/s 16 m/s.

Answer 1

Answer:

A- 3.5

Explanation: Edge

Answer 2

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

• Mass of the block, m = 2.5 kg
• Spring constant, k = 56 N/m
• Extension of the spring, x = 0.75 m

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

$K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5 \ m/s$

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here:  brainly.com/question/166559