What is the conservation of energy problem involving h and v?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Karhitect
Messages
10
Reaction score
0

Homework Statement


i attached the problem since its easier.


Homework Equations



im not sure. i know the equation for h has to be included. but I am not sure what other equations must be used.

The Attempt at a Solution



I honestly don't know how to begin. Any kind of help would be greatly appreciated!
 
Attachments
Physics news on Phys.org
i didnt think i had to use the potential spring energy equation because i need to find the velocity for when it comes back down...im really bad at this lol sorry
 
Karhitect said:

Homework Statement


i attached the problem since its easier.


Homework Equations



im not sure. i know the equation for h has to be included. but I am not sure what other equations must be used.

The Attempt at a Solution



I honestly don't know how to begin. Any kind of help would be greatly appreciated!

The terminology in the problem statement is confusing. When the problem refers to "strength", what it is really referring to is stiffness or spring constant. The term strength is usually reserved for when something breaks. You also apparently need to assume that the force the bow string exerts on the arrow is directly proportional to the draw displacement, such that when the draw displacement is zero, the force is zero, and when the displacement is d, the force is kd. So..., during the time that the arrow is in contact with the bow, what is the average force exerted on it by the bow? How much work is done on the arrow by the bow during the contact period? What does this translate into in terms of gravitational potential energy when the arrow reaches its maximum height. What is the upward velocity of the arrow when it reaches its maximum height?
 
would the equation be 1/2mv2=mgh

then replace h with the equation for h and isolate v??

im just not understanding this problem.
 
Last edited: