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

the maximum speed of the ball is 12.65 m/s

Step-by-step 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;

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

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