Tarzan Physics: Solving for Jane's Minimum Swinging Speed with Wind Resistance

[ad333]

I haven't done this be for so i could really use some help here.

Jane, Whos mass is 60kg, needs to swing to tarzan accrost a gap of 50m. There is a constant Horizontal force of 210N,caused by wind, opposing her. The length of the vine is 35m and the ange from were the vine is tied to jane is 50. what min speed must jane swing in order to get to tarzan and once there what speed must she use to get back with tarzan's 85kg weight added



I need help, i can't even figure out a basic equation to start with

 

[quasar987]

I believe you'll have to proceed with the energy method.

For a constant force, the work in traveling a distance x in the direction of the force is W = Fx. For us, W = 120x. Can you find the horizontal distance x from one point to the other? Then the minimum speed can be found by solving the equation Ti = W (where Ti is the initial kinetic energy, the final one being set to 0) for vi.

 

[ad333]

I believe you'll have to proceed with the energy method.

For a constant force, the work in traveling a distance x in the direction of the force is W = Fx. For us, W = 120x. Can you find the horizontal distance x from one point to the other? Then the minimum speed can be found by solving the equation Ti = W (where Ti is the initial kinetic energy, the final one being set to 0) for vi.

That does me no good i already know the distance, i need to know how to calculate the speed jane need to use

 

[Pyrrhus]

Consider This relationship

When jane swings to the other side

$$K_{o} + \Omega_{gravity_{o}} + W_{wind} = K + \Omega$$

This could be derived from Work-Kinetic Energy principle

$$W_{wind} - \Delta \Omega_{gravity} = \Delta K$$

 

[quasar987]

Now that I think about it, I believe it will be necessary to calculate the work by integrate over the curve using the angle as the variable of integration.

What is Omega of gravity cyclovenom?

 

[Pyrrhus]

Gravitational Potential Energy. I don't think that's necessary(spl?) quasar.