Work with counterweight question

[Halcyon32]

Hi, can someone explain this problem to me in a different way than the answer key:

"The box" is referring to a box on the lift plate. First, I'm confused about how the system even works. How can a box weighing 90 kg even be lifted by a counterweight weighing only 50 kg? Besides that, I don't really understand what the answer key is saying. It says that there is nothing to offset the counterweight once the box has been taken off the lift plate, but they didn't mention that being the case when the box was on the plate and it was basically like lifting a 40 kg box. Can someone please explain the reasoning here.

Also, I thought no work was performed when it was being done vertically, due to the cos90 being 0?

Here's the answer key explanation:

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[sazerac]

When you have a 90kg mass on the ground, with a 50kg counterweight in the air attached via a pulley on the ceiling, you only have to exert enough force to lift an equivalent 40kg mass. The counterweight doesn't do all the work for you, but it does assist you.

Once the 90kg mass is lifted, and disconnected from the system, you still have a 50kg counterweight sitting on the ground all by itself. It takes more force to lift that lonely 50kg counterweight than it did to do the lifting described in the previous paragraph.

Theta is defined as the angle between the direction of force and the direction of motion. In this case they both point in the same direction (up), so theta is zero, and cos(theta) is 1, and the term drops out.

 

[Halcyon32]

When you have a 90kg mass on the ground, with a 50kg counterweight in the air attached via a pulley on the ceiling, you only have to exert enough force to lift an equivalent 40kg mass. The counterweight doesn't do all the work for you, but it does assist you.

Once the 90kg mass is lifted, and disconnected from the system, you still have a 50kg counterweight sitting on the ground all by itself. It takes more force to lift that lonely 50kg counterweight than it did to do the lifting described in the previous paragraph.

Theta is defined as the angle between the direction of force and the direction of motion. In this case they both point in the same direction (up), so theta is zero, and cos(theta) is 1, and the term drops out.

Ok, that was a lot simpler than I made it out to be. Idk why that was so hard for me to understand. Thanks a lot!

 

[sazerac]

Ok, that was a lot simpler than I made it out to be. Idk why that was so hard for me to understand. Thanks a lot!

A lot of the MCAT passages are just familiar concepts presented in unfamiliar situations. Recognizing them is a skill in and of itself, and a skill that the MCAT rewards.

Sorry you got tripped up by the confusing problem setup this time. Better luck next time!!

 

[Halcyon32]

A lot of the MCAT passages are just familiar concepts presented in unfamiliar situations. Recognizing them is a skill in and of itself, and a skill that the MCAT rewards.

Sorry you got tripped up by the confusing problem setup this time. Better luck next time!!

Yeah, I'm coming to realize that. Hopefully doing tons of passages helps with that. Thanks again!