Finding work done by object along circular helix: Line Integral

yolanda
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Homework Statement


An object weighing 1.2 pounds travels along a helix given by x=cost, y=sint, z=4t, 0<=t<=8pi. Find the work done by the object.

Let's keep this in ft.

Homework Equations


g=32.174 ft/s2
f=m*g
f=w*d

The Attempt at a Solution



r(t)=cos(t)i+sin(t)j+4(t)k

I know I need an F function, but I'm not sure how to find it.

w=\int F(dot)dr
w=\intF(r(t)) (dot) r'(t) dt

Any help would be appreciated, thanks in advance!
 
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You are working against gravity, which is directed downward. So figure out the force f of gravity on your object and use

\vec F = \langle 0,0,-f\rangle

This is the work done by the force, opposite the work done by the object.
 
Thank you for your response. This may sound a bit stupid, but I want to make sure I'm getting the right force here...

Weight is the mass of an object with a force acting upon it. I'm not sure how much the equations f=ma and weight=m*gravity "overlap", if you will. Can I just say 1.2=mass*gravity? That would mean that the force is 1.2 pounds... do you see what I'm confused about here?

Thanks again for your help!
 
1.2 pounds is a unit of force in the English system. And work is measured in foot-pounds which gives the right units for w = fd in the English system.
 
yolanda said:
Weight is the mass of an object with a force acting upon it.
Weight is a force. The formula F = ma gives the force F acting on an object of mass m under an acceleration a.

In the SI system, a mass of 1 kg at the surface of the Earth exerts a force downward of 1kg * 9.8 m/s^2 = 1 Nt. In the English system, a mass of 1 slug at the surface of the Earth exerts a force downward of 1 slug* 32 ft/sec^2 = 32 lb.
yolanda said:
I'm not sure how much the equations f=ma and weight=m*gravity "overlap", if you will. Can I just say 1.2=mass*gravity? That would mean that the force is 1.2 pounds... do you see what I'm confused about here?

They overlap considerably, but the F = ma formula applies more generally for any kind of acceleration, not just that due to gravity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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