A spring with a force constant of 5400 N/m and a rest length of 3.5 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m) ? (Assume the rock is launched from ground height.) m

5.51 m

Explanation:

From the question,

The energy used to stretch the spring = the potential energy of the rock.

(1/2) ke² = mgh ... Equation 1

Where k = spring constant, e = extension/compression, m = mass of the rock, g = acceleration due to gravity, h = height of the rock above the ground

make h the subject of the equation.

h = ke²/2mg ... equation 2

Given: k = 5400 N/m, e = 1 m, m = 48 kg.

Constant: g = 9.8 m/s²

Substitute into equation 2

h = 5400 (1²) / (2*48*9.8)

h = 5400/940.8

h = 5.51 m.

Hence the height of the rock = 5.51 m