Question

A block of mass m=0.750 kg is fastened to an unstrained horizontal spring whose spring constant is k=82.0 N/m. The block is given a displacement of +0.120m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. C) What is the maximum speed of the block? Group of answer choices

Answer 1

Answer:

13.12 m/s

Explanation:

According to the law of energy conservation, the maximum speed of the block occurs when all the spring potential energy is converted to block kinetic energy. That is:

$E_s = E_k$

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

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

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

we can substitute m = 0.75 kg, k = 82 N/m, and x = 0.12m

$v = 0.12\frac{82}{0.75} = 13.12 m/s$