Work done by gravity by Wonder Woman

[venceslau]

Work done against gravity by Wonder Woman

Homework Statement

Wonder Woman, whose mass is 52 kg, is holding onto the free end of a 13.0 m rope, the other end of which is fixed to a tree limb above. She is able to get the rope in motion so that she can reach a ledge when the rope makes a 60 degree angle with the downward vertical axis. How much work does Wonder Woman do against the force of gravity?

Homework Equations

W=mgdcos(θ) ...maybe?

The Attempt at a Solution

Honestly, I'm having a hard time even understanding the wording of the problem itself (ESL). Are they telling me she swung over to a ledge, and want me to calculate the work of gravity when she's by that ledge? I don't get it...some insight on what is being asked would be greatly appreciated.

Thanks in advance,

Filipe

 

[PeterO]

Homework Statement

Wonder Woman, whose mass is 52 kg, is holding onto the free end of a 13.0 m rope, the other end of which is fixed to a tree limb above. She is able to get the rope in motion so that she can reach a ledge when the rope makes a 60 degree angle with the downward vertical axis. How much work does Wonder Woman do against the force of gravity?

Homework Equations

W=mgdcos(θ) ...maybe?

The Attempt at a Solution

Honestly, I'm having a hard time even understanding the wording of the problem itself (ESL). Are they telling me she swung over to a ledge, and want me to calculate the work of gravity when she's by that ledge? I don't get it...some insight on what is being asked would be greatly appreciated.

Thanks in advance,

Filipe

I suspect the work done will simply equal the change in Gravitational potential energy compared to hanging vertically, and being on the ledge.

 

[venceslau]

I suspect the work done will simply equal the change in Gravitational potential energy compared to hanging vertically, and being on the ledge.

Ohh...I feel so dumb now. lol.
So, will it be just the change in potential energy for Δh?

[STRIKE]So final equation is W=m*g*Δh=m*g*(30-30*cos(θ)) ?[/STRIKE]
So final equation is W=m*g*Δh=m*g*(13-13*cos(θ)) ?

Thanks

 

[PeterO]

Ohh...I feel so dumb now. lol.
So, will it be just the change in potential energy for Δh?

So final equation is W=m*g*Δh=m*g*(30-30*cos(θ)) ?

Thanks

I think the figure [rope length] in the original was 13 not 30?
I appreciate the ESL difficulty with written questions.

 

[venceslau]

I think the figure [rope length] in the original was 13 not 30?
I appreciate the ESL difficulty with written questions.

Thanks for pointing that out! Fixed.
I am the worst for making dumb mistakes like that.

I appreciate your quick reply, I can now finish my homework. :)