Calculating Work for a Ballista

[Arbitrary]

Homework Statement

A ballista is essentially a very large bow and arrow, except that it fires 3-m long, 2-kg arrows. The arrows are propelled by the stretched bowstring and bow, which behave as if they are two nonlinear springs that each behave according to the equation
F = -kx^1.5, where the force constant k = 500 N·m-1.5 and equilibrium length l = 0.5 m. The ballista is cocked, so that the distances x and y are 1.0 m and 1.3 m, respectively. A constant 10-N frictional force opposes all motion of the arrow. The arrow is shot straight upwards.

How much work was required to attach the bowstring to the bow?

How much work was required to cock the ballista?

Homework Equations

dW = F dx

The Attempt at a Solution

I just used the equation above, set F to -kx^1.5 and integrated from x=0 to x=0.5.

I did the same thing for the second question, except that I integrated from x=0 to x=0.8. I'm not sure what exactly I did wrong...

 

[learningphysics]

Hmmm... do they give a picture? I'm finding it hard to understand the problem exactly. What do the x and y distances mean? "x and y are 1.0 m and 1.3 m"

 

[m2003]

I would like to provide an alternative solution to the given problem. First, let's define some variables for easier understanding. Let x be the distance from the equilibrium point of the bowstring, y be the distance from the equilibrium point of the bow, and L be the total length of the bowstring (L=x+y).

To attach the bowstring to the bow, we need to stretch the bowstring from its equilibrium point to a length of L=1.3 m. This can be done by applying a force F at a distance of x=1.0 m from the equilibrium point of the bowstring. This force F can be calculated using Hooke's Law, F=kx, where k is the force constant given in the problem. Therefore, the work required to attach the bowstring to the bow can be calculated as follows:

W = ∫F dx = ∫kx dx = ½ kx^2 = ½ (500 N·m-1.5)(1.3 m)^2 = 337.5 J

To cock the ballista, we need to stretch the bow from its equilibrium length of 0.5 m to a length of L=1.3 m. This can be done by applying a force F at a distance of x=0.8 m from the equilibrium point of the bow. Again, using Hooke's Law, we can calculate the required force as:

F = kx = (500 N·m-1.5)(0.8 m) = 200 N

The work required to cock the ballista can then be calculated as:

W = ∫F dx = ∫kx dx = ½ kx^2 = ½ (500 N·m-1.5)(0.8 m)^2 = 128 J

It is important to note that the given equation, F = -kx^1.5, is not a linear equation and cannot be directly integrated to calculate work. Instead, we can use the general formula for work, W = ∫F dx, and use the appropriate values for x and F to calculate the work required.