A archer puts a .3kg arrow into the bowstring. An average force of 201N sis exerted to draw the bowstring 1.3m.
a) if there is no friction, what velocity does the arrow have when it leaves the bow?
When i worked it out using two different ways, the answer is conpletely different. They are supposed to be the same.
here's the first way:
To find the energy store in the bowstring
W=Fx
W= (201N) (1.3m)
W= 261.3J
W=KE
W= 1/2mv^2
261.3J = 1/2 (.3kg)v^2
V= 41.73m/s
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here is another way that i used.
I find the spring constant first
F=kx
201N = k(1.3m)
k= 154.62 N/m
EPE = 1/2kx^2
EPE = 1/2(154.62N/m)(1.3m)^2
EPE = 130.65J
EPE=KE
EPE=1/2mv^2
130.65J = 1/2(.3kg)v^2
v= 29.5m/s
i don't know which way is correct, this really drives me crazy. If the first way is correct, then why not the second way? and if the second way is correct, then why not the first way?
can somebody please explain??? i really need help.
Update:first of all, thank you for answering.
and second, you didn't really answer my question. i doubt if you read the last paragraph.
Answers & Comments
Verified answer
The mistake you made in your first method of solving the problem was that you assumed that all the work that was done by pulling the bow back went into potential energy. I will try to show why this assumption is not true.
F(x) = -dU/dx (U = potential energy)
If F(x) = kx, then
kx = -dU/dx
To find potential energy, you integrate both sides to get:
U(x) = 1/2kx^2
As you can see, Fx = (kx)(x) = kx^2is not equal to 1/2kx^2. Only half of the work you put into getting the bow in position actually goes towards the potential energy of the system.
Your second method of solving the problem is correct. I hope this helps and I hope I explained it correctly.
Is this question based upon the fact that you need to find a solution? If this is from a text book there can be no answer. There are too many variables and no constants. A question of stupidity regardless of how many years you have in a professor ship. If you are a student I extend my sincere sympathy.