Question

3. A 40-gram ball of clay is dropped from a height, h, above a cup which is attached to a spring of spring force constant, k, of 25 N/m. The ball lands in the cup, and the spring compresses a distance, x. If x is 0.506 m, then what is the maximum speed of the ball?

Answer 1

Answer:

the maximum speed of the ball is 12.65 m/s

Explanation:

Given;

mass of the ball, m = 40 g = 0.04 kg

spring constant, k = 25 N/m

Apply the principle of conservation of energy;

The Elastic potential energy of the spring will be converted into Kinetic of the ball;

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\ kx^2 = mv^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(25)(0.506)^2}{0.04}} \\\\v = \sqrt{160.0225} \\\\v = 12.65 \ m/s$

Therefore, the maximum speed of the ball is 12.65 m/s