Question

3) An archer pulls back the arrow in his bow a distance of 0.5 m against an average force of 200 N. a) How much work is done? b) What are the energy changes here?

Answer 1

$\underline{\boxed{\sf{Given\: : }}}$

Distance (s) = 0.5 m

Force (F) = 200 N

$\underline{\boxed{\sf{To\:Find\: :}}}$

Potential Energy Gained

$\underline{\boxed{\sf{Solution\: : }}}$

We are given Force with which the archer pulls an arrow from her bow, and also the distance stretching is 0.5 m.

Here, work done is stored as Potential Energy.

$\implies Work\:Done=Potential\:Energy\\\\\implies\therefore P.E=F.s$

Here,

• P.E = Potential Energy
• F= Force with which arrow is pulled from now
• s=distance

$\implies P.E =F.s\\\\\implies P.E=200×0.5\\\\\implies P.E = 100 \:joules\\\\\\\\\underline{\boxed{\sf{P.E\: gained (i.e\:work\:done)\:is\:100\:joules}}}$

Answer 2

Answer:

h

Explanation: