Aspring with a force constant of 5100 n/m and a rest length of 2.6 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 how fast is it going (in m/s) when it hits the ground? m/s

First let’s find the total force exerted on the rock.
Fs=1/2kx^2= 1/2*5100*(1.6)^2=6528N
Then setting it equal to U=mgh and solving for h we get h=6528N/(9.8m/s^2*48kg)= 13.9m

Then using K=1/2mv^2 we can solve for the final velocity of the rock.
(6528N*2)/48kg then taking the sqrt of that we get Vf= 16.5m/s

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf,  ℰ,  to a battery's terminal voltage, v, current, i, and internal resistance, r, is    ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.