Question

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

Answer 1

Answer:

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