Question

A spring with a force constant of 4900 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 42 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.)

b.

How fast is it going (in m/s) when it hits the ground?

Answer 1

Part A,

Using energy conservation between initial and final position of rock:

KEi + PEi = KEf + PEf

KEi = 0, since initially rock was at rest,

PEi = energy stored in spring = (1/2)*k*dx^2 (Initial GPE is zero, since given that rock is launced from ground height)

KEf = 0, since KE at max height will be zero

PEf = m*g*hf

So,

0 + (1/2)*k*dx^2 = 0 + m*g*hf

hf = k*dx^2/(2*m*g)

k = 4900 N/m and dx = compression of spring = 3.5 m - 1.0 m = 2.5 m

m = mass of rock = 42 kg

So,

hf = 4900*2.5^2/(2*42*9.8)

hf = 37.2 m = max height above the ground

Part B.

Now after reaching max height, rock will start traveling downward with zero velocity, So using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*g*hf

Vf = sqrt (0^2 + 2*9.8*37.2)

Vf = 27.0 m/s = final speed of rock when it hits the ground

Let me know if you've any query.