Hey!

Is there anyone taking physics 1B03 that has completed the CAPA assignment due tonight?
I've got just 3 questions left.. (3 parts to the same q) and I can't seem to understand why my method of approach is wrong.
If there's someone that could help me out, please message me! Thank you

You could always post the questions and those of us here who know how to solve the problem who are not in physics 1b03, yet still know physics, could run you through how to solve it without giving you a final answer (as that would probably constitute academic dishonesty)

Alright, here's the question:

A broken clothes drier with a mass of 121 kg is hurled from a cliff 13.9 m high with an initial velocity of 13.4 m/s, directed at an angle of 39.8° above the horizontal. How much work is done on the drier by its own weight during its fall from the cliff to the ground? What change in the gravitational potential energy of the clothes drier takes place during the fall? If the gravitational potential energy at the top of the cliff is assumed to be zero, what is its value when the clothes drier crashes into the ground?

And yes, I don't want someone to give me the final answers. I'd like if someone can work me through the problem so I could see where I went wrong and derive the answers on my own.

Lemme see if I can remember how to do this from my grade 12 physics (which covered more than 1L03 has covered thus far...)

#1) Work=force*distance Since it says work done by weight does it not only consider the y axis? (13.9m) If that's wrong you probably have to consider how high above the trajectory the throw puts the drier above the cliff and add it to the 13.9 m.

#2 & #3)
for 3 split (delta)Ep into Ep(cliff)-Ep(ground) where Ep(cliff)=0 J

Originally Posted by ShesTheMan

Alright, here's the question:

A broken clothes drier with a mass of 121 kg is hurled from a cliff 13.9 m high with an initial velocity of 13.4 m/s, directed at an angle of 39.8° above the horizontal. How much work is done on the drier by its own weight during its fall from the cliff to the ground? What change in the gravitational potential energy of the clothes drier takes place during the fall? If the gravitational potential energy at the top of the cliff is assumed to be zero, what is its value when the clothes drier crashes into the ground?

And yes, I don't want someone to give me the final answers. I'd like if someone can work me through the problem so I could see where I went wrong and derive the answers on my own.

A) Remember, Work = Force * Distance. So, what's the Force exerted on the drier as it's falling and what was the distance covered from the moment the drier was hurled? Remember, it's fall from the cliff to the ground also includes the upward distance the drier travelled.

B) I don't really know what it's asking for... The ambiguity of the question makes me think it's just asking if the change in gravitational potential is positive or negative... However it could also mean, what is the biggest change in energy.

C) This is just using another formula, which is the gravitation potential energy. It's been a while, but I think it's mgh.

Unrelated: Clothes driers are recyclable, do not throw them off cliffs. THIS IS A BAD THING TO DO.

For #1, I calculated the additional distance it travels upwards and multiplied it by mg because its asking for the work done by the weight. But that's the same thing has potential gravitational energy and apparently, that's incorrect.

Originally Posted by ShesTheMan

For #1, I calculated the additional distance it travels upwards and multiplied it by mg because its asking for the work done by the weight. But that's the same thing has potential gravitational energy and apparently, that's incorrect.

Wait, did you just multiply only the upwards distance by mg? Shouldn't it be (upwards distance+13.9 m)*mg?

Originally Posted by REPLEKIA/.

Wait, did you just multiply only the upwards distance by mg? Shouldn't it be (upwards distance+13.9 m)*mg?

I believe it should actually be (2*(Upwards Distance)+13.9)*mg...

Noo. my bad.
I calculated the additional distance it travelled upwards, added that to the 13.9 and then multiplied the entire distance by mg
And it's still wrong :(

Maybe my calculations were wrong? Is someone would like to double check.

I did (13.4sin39.8)^2 / (2*9.81) = 3.75 m (the additional distance it travelled upwards)
and then W =mgd = (121)(9.81)(3.75+13.9 ) = 20950 J

Originally Posted by jhan523

I believe it should actually be (2*(Upwards Distance)+13.9)*mg...

Sorry I'm having trouble understanding... why would it be 2*upwards distance?

Originally Posted by Physics forum elsewhere on the internet

The concept of work arises only when there is a force acting on a body. In case of a projectile no force acts in the X direction as the X component of velocity remains constant. But the force of gravity acts in the Y direction. therefore work would be mg*(component of displacement in the Y direction of the projectile).

Therefore work should be 121 kg*9.8 m/s^2*13.9 m

If that doesn't work than maybe CAPA is getting mad over how you enter the answer.

^ No, that worked.
Thanks a lot!

Originally Posted by ShesTheMan

Sorry I'm having trouble understanding... why would it be 2*upwards distance?

Oh ok, sorry about that. Work is not Force*Distance, it's Force*Displacement >.<

Originally Posted by ShesTheMan

^ No, that worked.
Thanks a lot!

did you get B and C as well?

How did you work those out?