Why Does Work Equal Zero in Some Situations?

[dwdoyle8854]

Homework Statement

Im having difficulties with this on several problems. I understand work to be zero when there is either no displacement, no forces acting parallel to object or if friction and air resistance is not considered.

Here is a simpler problem where i get the right answer if i set W = 0 but where i don't see WHY work done is zero:

Harry is pulling a 25 pound (max pull) bow. The 1 oz. arrow is said to attain a speed of 140 feet/sec after release of the bowstring. If harry pulled the arrow back 18 inches prior to release, CONFIRM the arrow does leave the bow at 140 ft/s.

Homework Equations

Potential Energy elastic = U = 1/2kx^2 where k = F/x
Kinetic Energy = K = 1/2mv^2
Work = delta K + delta U

The Attempt at a Solution

I set origin at where the string is at maximum pull
so delta x is = 1.5 feet = 1.5 - 0
U initial = 1/2kx^2
k = F/x = ~16.667
U final = 0
K initial = 0
K final = 1/2mv^2


if work = force * delta x, then 25 * 1.5 = 37.5 Joules is my work.

with W= delta K + delta U

W = 37.5 J = 1/2mv^2 - 1/2kx
(75 + kx)/m=v^2
v=240 ft/s

but if i work this with w = 0 = 1/2mv^2 - 1/2kx
v is verified at 140 ft/s

I can see the work done on the string is zero, as it is pulled from an initial point and returns to the point, but how and why is work done on the arrow zero, if there is a displacement and a force of 25 lbs acting on it? Please help!

 

[vkash]

Im having difficulties with this on several problems. I understand work to be zero when there is either no displacement, no forces acting parallel to object or if friction and air resistance is not considered.

dW=F.ds (. represent dot/scalar product)
try to understand this formula. this small 3 variable formula can tell you when work is zero.
case(1) when F=0.
case(2) when ds=0
case(3) when F.ds=0 but neither F not ds is zero. In other words F(vector) is perpendicular to ds(vector).
I doesn't understand your question, But i think i have answered you when work is zero which is your main question.

 

[dwdoyle8854]

Why is the work zero in this application?

W = Force * displacement

Force is 25, there are non conservative forces working so E is not constant, there is a displacement of 1.5 feet, and all the forces acting are parallel to the arrow.

?

 

[Pengwuino]

You have to consider this in steps. As you pull back the bow, nothing is actually done to the arrow! Think about it, if you pull back the string, you could simply pull back the string without an arrow and nothing would really change. So the work done on the bow is W_{bow} = \Delta U = .5k\Delta x^2, just a change in potential. The work done on the arrow is 0.

Upon releasing, the work done by the string is equal to the work done on the arrow. In other words

W_{bow} = -W_{arrow}

where the negative sign is because one is work done to an object and one is work done on an object. This is because of conservation of energy ( W_{bow} + W_{arrow} = 0 if it's a closed system) The work done on the bow, when released, is -.5k\Delta x^2 (the negative because the bow actually does work and thus, loses energy in the form of potential energy). The work done on the arrow, when the bow is released, ends up being .5mv^2.

 

[gneill]

Homework Statement

Im having difficulties with this on several problems. I understand work to be zero when there is either no displacement, no forces acting parallel to object or if friction and air resistance is not considered.

Here is a simpler problem where i get the right answer if i set W = 0 but where i don't see WHY work done is zero:

Harry is pulling a 25 pound (max pull) bow. The 1 oz. arrow is said to attain a speed of 140 feet/sec after release of the bowstring. If harry pulled the arrow back 18 inches prior to release, CONFIRM the arrow does leave the bow at 140 ft/s.

Homework Equations

Potential Energy elastic = U = 1/2kx^2 where k = F/x
Kinetic Energy = K = 1/2mv^2
Work = delta K + delta U

The Attempt at a Solution

I set origin at where the string is at maximum pull
so delta x is = 1.5 feet = 1.5 - 0
U initial = 1/2kx^2
k = F/x = ~16.667
U final = 0
K initial = 0
K final = 1/2mv^2


if work = force * delta x, then 25 * 1.5 = 37.5 Joules is my work.

That's not right; the bow acts as a spring, so the force will vary with the displacement of the string. I suggest you determine how much potential energy the bow holds at its maximum pull. That energy will be delivered to the arrow...

 

[vkash]

Harry is pulling a 25 pound (max pull) bow. The 1 oz. arrow is said to attain a speed of 140 feet/sec after release of the bowstring. If harry pulled the arrow back 18 inches prior to release, CONFIRM the arrow does leave the bow at 140 ft/s.

Homework Equations

Potential Energy elastic = U = 1/2kx^2 where k = F/x
Kinetic Energy = K = 1/2mv^2
Work = delta K + delta U

The Attempt at a Solution

I set origin at where the string is at maximum pull
so delta x is = 1.5 feet = 1.5 - 0[/color]
U initial = 1/2kx^2
k = F/x = ~16.667
U final = 0
K initial = 0
K final = 1/2mv^2if work = force * delta x, then 25 * 1.5 = 37.5 Joules is my work.

with W= delta K + delta U

W = 37.5 J = 1/2mv^2 - 1/2kx
(75 + kx)/m=v^2
v=240 ft/s

but if i work this with w = 0 = 1/2mv^2 - 1/2kx
v is verified at 140 ft/s

I can see the work done on the string is zero, as it is pulled from an initial point and returns to the point, but how and why is work done on the arrow zero, if there is a displacement and a force of 25 lbs acting on it? Please help!

got the question..
what is delta x in the formula?
that is change in length of the string. here your friend() harry pulled rope 18inches back so the change in length of string will not equal to 18 inches.
You need to make full diagram of step when bow is stretched by full strength and at the moment when bow is not stretched. You can see the change in length of string. then apply some simple trigonometry. you will got sin(θ) or coos(θ) (based on your assumption that what is θ) in terms of tension in string. after that equate with the change in kinetic energy in bow with PE of string. i mean all the energy of arrow (after getting140f/s) is equal to potential energy of bow at this moment. after this you will got change in string length. rest of data can be attained from delta x.
But why are you doing all this. the work done by harry is equal to change in KE of bow. since harry uses his muscular power to give velocity to bow. Most of all, i don't think that the work done by harry is zero.

 

[dwdoyle8854]

You have to consider this in steps. As you pull back the bow, nothing is actually done to the arrow! Think about it, if you pull back the string, you could simply pull back the string without an arrow and nothing would really change. So the work done on the bow is W_{bow} = \Delta U = .5k\Delta x^2, just a change in potential. The work done on the arrow is 0.

Upon releasing, the work done by the string is equal to the work done on the arrow. In other words

W_{bow} = -W_{arrow}

where the negative sign is because one is work done to an object and one is work done on an object. This is because of conservation of energy ( W_{bow} + W_{arrow} = 0 if it's a closed system) The work done on the bow, when released, is -.5k\Delta x^2 (the negative because the bow actually does work and thus, loses energy in the form of potential energy). The work done on the arrow, when the bow is released, ends up being .5mv^2.

if energy lost = energy gained

why can't I merely set ΔU = ΔK ?