Question

An archer pulls back her bowstring loaded with an arrow thats .22kg. It leaves the arrow at 220 kmph, calculate work done.

Answer 1

By Work Energy Theorem,
Work Done = Change in Kinetic Energy
KE(Initial)=0
KE(Final)=\[\frac{ mv^{2} }{ 2 }\]
\[W=0-\frac{ mv^{2} }{ 2 }\]
\[W=-\frac{ mv^{2} }{ 2 }\]
Mass m of the bow is given, velocity v of the bow is given, make sure to convert Km/hr to m/s
Use the factor of 5/18
\[v(m/s)=v(km/hr)*\frac{ 5 }{ 18 }\]

Answer 2

I had it! I was using 220 instead of 61.1 though >.< Woops!