Calculating Arrow Speed from Energy Transfer: Archer's Dilemma [SOLVED]

[zvee_y]

[SOLVED] energy transfer?

Homework Statement

An archer puts a 0.27 kg arrow to thebowstring. An average force of 184.5 N ix exerted to draw the string back 1.3m. Assuming no frictional loss, with what speed does the arrow leave the bow?

The Attempt at a Solution

So i use the equation (1/2)mv^2 = (1/2)kx^2 = v= square root of (141.923*1.3^2)/(0.27) = 29.8 . But the answer is wrong. Please help me out! Thanks a lots

 

[Dick]

They are giving you average force and distance. There is a much more basic solution than to try to assume it is some kind of spring. Work is equal to average force times distance, isn't it?

 

[zvee_y]

Oh yeah. What if the arrow is shot striaght up, how high does it rise then?
Is it right if we use equation mgh=(1/2)kx^2
find k by dividing the average force by the distance the string draw back?

 

[Dick]

You haven't answered the first question yet. You only use (1/2)kx^2 in SPRING problems. This isn't posed as a spring problem. You don't use it.

 

[zvee_y]

Yes I got the 1st one by using Work=Fd and then transfer it to kinetic energy=(1/2)mv^2 . Find v from that. Because this question is from spring's homeworks. So i just try to see how do we find the height of arrow goes if we shot it vertically?

 

[Dick]

You find the height of the arrow by equating kinetic and mgh potential energy, just like usual. Just because the problem is in the spring section doesn't mean it's directly related to springs. Apparently.