Question

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming that all the energy goes into the arrow, with what speed does the arrow leave the bow? (41.7 m/s)b. If the arrow is shot straight up, how high does it rise? (88.9 m)

Answer 1

Step-by-step explanation:

Given

mass of archer
`m=0.3\ kg`

Average force
`F_(avg)=201\ N`

extension in arrow
`x=1.3\ m`

Work done to stretch the bow with arrow

`W=F\cdot x`

`W=201* 1.3=261.3\ m`

This work done is converted into kinetic Energy of arrow

`W=(1)/(2)mv^2`

where v= velocity of arrow

`261.3=(1)/(2)* 0.3* v^2`

`v=√(1742)`

`v=41.73\ m/s`

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

`W=mgh`

`261.3=0.3* 9.8* h`

`h=(261.3)/(0.3* 9.8)`

`h=88.87\ m`