Question

A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30-kg ball is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the uncompressed spring was the block dropped

Answer 1

Answer:

h = 1.91 m

Explanation:

From the law of conservation of energy, we know that:

$Potential\ Energy\ of\ Ball = Elastic\ Petential\ Energy\ of\ Spring\\mgh = \frac{1}{2}kx^2$

where,

m = mass of ball = 0.3 kg

g = acceleration due to gravity = 9.81 m/s²

h = height of ball = ?

k = spring constant = 450 N/m

x = compressed length = 2.5 cm = 0.025 m

Therefore,

$(0.3\ kg)(9.81\ m/s^2)h = \frac{1}{2} (450\ N/m)(0.025\ m)^2\\\\h = \frac{(450\ N/m)(0.025\ m)^}{(2)(0.3\ kg)(9.81\ m/s^2)}$

h = 1.91 m