Calculating work done using line integrals

In summary: W=\int_0^{10} 50x'(t) \cdot x'(t)dt=\int_0^{10} 50\left\|x'(t)\right\|^2dt=\int_0^{10} 50(325+100)dt=21500## Joules.In summary, Sisyphus exerts a force of 50 Ib tangent to his path as he pushes the boulder up a 100-ft tall spiral staircase surrounding a cylindrical castle tower. Using the parametric equation of his path, it was found that he performs a total work of 21500 Joules.
  • #1
toforfiltum
341
4

Homework Statement


Sisyphus is pushing a boulder up a 100-ft tall spiral staircase surrounding a cylindrical castle tower.

a) Suppose Sisyphus's path is described parametrically as $$x(t)=(5\cos3t, 5\sin3t, 10t)$$, $$\space 0\leq t\leq10$$.
If he exerts a force with constant magnitude of 50 Ib tangent to his path, find the work Sisyphus does in pushing the boulder up to the top of the tower.

b) Just as Sisyphus reaches the top of the tower, he sneezes and the boulder slides all the way to the bottom. If the boulder weighs 75 Ib, how much work is done by gravity when the boulder reaches the bottom?

Homework Equations

The Attempt at a Solution


OK, I'm stuck at a), so I would just type out what I have done so far.
I know that I must find ##x'(t)##, so ##x'(t)=(-15\sin3t, 15\cos3t, 10)##. However, this is where I'm stuck. Since usually, force is given in its ##xyz## components, it is easy for me to just do the dot product. But here, it just states that the force is 50 tangent to the path. How do I find the force in the ##xyz## components? Any hints?

Thanks!
 
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  • #2
toforfiltum said:

Homework Statement


Sisyphus is pushing a boulder up a 100-ft tall spiral staircase surrounding a cylindrical castle tower.

a) Suppose Sisyphus's path is described parametrically as $$x(t)=(5\cos3t, 5\sin3t, 10t)$$, $$\space 0\leq t\leq10$$.
If he exerts a force with constant magnitude of 50 Ib tangent to his path, find the work Sisyphus does in pushing the boulder up to the top of the tower.

b) Just as Sisyphus reaches the top of the tower, he sneezes and the boulder slides all the way to the bottom. If the boulder weighs 75 Ib, how much work is done by gravity when the boulder reaches the bottom?

Homework Equations

The Attempt at a Solution


OK, I'm stuck at a), so I would just type out what I have done so far.
I know that I must find ##x'(t)##, so ##x'(t)=(-15\sin3t, 15\cos3t, 10)##. However, this is where I'm stuck. Since usually, force is given in its ##xyz## components, it is easy for me to just do the dot product. But here, it just states that the force is 50 tangent to the path. How do I find the force in the ##xyz## components? Any hints?

Thanks!

The total work Sisyphus performs is ##W = \int \vec{F} \cdot d\vec{s}##, where ##d\vec{s}## is the arc-length along the tangent. You were told that ##\vec{F}\, \| \,d\vec{s}## at all points.
 
  • #3
Ray Vickson said:
The total work Sisyphus performs is ##W = \int \vec{F} \cdot d\vec{s}##, where ##d\vec{s}## is the arc-length along the tangent. You were told that ##\vec{F}\, \| \,d\vec{s}## at all points.
Yes, I think I've got it. ##F## is ##50 \frac {x'(t)}{\left\|x'(t)\right\|}##
 

Related to Calculating work done using line integrals

1. What is a line integral?

A line integral is a mathematical concept used to calculate the work done by a vector field along a specific curve or path. It takes into account the magnitude and direction of the vector field and the displacement of the curve.

2. How is work calculated using line integrals?

To calculate work using line integrals, you first need to determine the vector field and the path along which the work is being done. Then, you integrate the dot product of the vector field and the tangent vector of the path along the curve.

3. What are some real-life applications of line integrals?

Line integrals have many practical applications, such as calculating the work done by a force along a certain path, finding the mass of a wire by measuring its length and density, and calculating the circulation of a fluid flow around a closed loop.

4. Can line integrals be negative?

Yes, line integrals can be negative. This typically occurs when the vector field and the path have opposite directions or when the work is being done in the opposite direction of the vector field.

5. Are there any limitations to using line integrals to calculate work?

Line integrals can only be used to calculate work in vector fields that are conservative, meaning that the work done is independent of the path taken. They also cannot be used to calculate work in non-conservative fields, such as those with a changing magnetic field or a frictional force.

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