Hi,
Why don't we have to use the angle in this question? (W=Fd x cos(theta))? Since you're pulling with an angled force, wouldn't it be W=Fd x cos(60)?
Thank you,
AAMC Physics Qpack #74

 Posts: 159
 Joined: Sat Mar 30, 2019 8:39 pm
Re: AAMC Physics Qpack #74
Work is a state function. The work done to lift an object is independent of whether you pull on that string at 1 degree or 90 degrees, whether you do it slowly or quickly. The object itself is rising at 90 degrees to the horizontal plane, and the force acting on it is the force of gravity: So 4 x 10 x 5 = 200.
This is an excellent demonstration problem for why solving for work is much easier than anything else: You just need to know the force and distance of the object work was done on and nothing else. You have no reason to care how it was done! Keep this in mind for slope & spring problems, in comes in really handy.
Edit: A slightly more comprehensive way of looking at this is also that the work done must be equal to the change in potential energy of that mass M.
This is an excellent demonstration problem for why solving for work is much easier than anything else: You just need to know the force and distance of the object work was done on and nothing else. You have no reason to care how it was done! Keep this in mind for slope & spring problems, in comes in really handy.
Edit: A slightly more comprehensive way of looking at this is also that the work done must be equal to the change in potential energy of that mass M.
Re: AAMC Physics Qpack #74
Thank you. This cleared it up a bit, but I'm still a little confused.
When we solved for force using F=mg, that gives us the force (F) shown in the diagram, correct? Wouldn't be have to find the vertical component of force & use that in the W=Fd equation since the weight would be moving in the vertical direction? Similarly to the example in the physics review book (pg. 47) when they used the horizontal component of force in the equation since the box was moving horizontally?
Thanks again!
When we solved for force using F=mg, that gives us the force (F) shown in the diagram, correct? Wouldn't be have to find the vertical component of force & use that in the W=Fd equation since the weight would be moving in the vertical direction? Similarly to the example in the physics review book (pg. 47) when they used the horizontal component of force in the equation since the box was moving horizontally?
Thanks again!

 Posts: 159
 Joined: Sat Mar 30, 2019 8:39 pm
Re: AAMC Physics Qpack #74
The force shown in the diagram is a red herring: You just don't care about it in any way. It would be impossible to solve this way since you don't know how far the rope on the pulley was pulled.
A mass M moves from height = 0 to height = 5 m, and it does so straight vertically (as shown in the drawing with the pulleys). How you pull on it just doesn't matter. The force acting on it is that of gravity. Or alternatively you can just frame this in terms of potential energy: PE_1 = mgh = 4*0*10 = 0 and PE_2 = mgh = 4*5*10 = 200
Same as if I had a frictionless ramp of any angle. If I push something from the bottom to the ramp to a height of 5 meters, the work done is the same, regardless of the slope of the ramp.
Edit: I was silly, the rope itself would be pulled 5 meters. You could if you wanted to solve for the force first, and then the work done, but you would have to keep straight that there is now another actor not on the diagram doing the work and you're really calculating the work done by him or her. And for all that effort, you will find the two cos(60) terms canceling each other and your answer being identical. Identifying the efficient way of solving problems is a big part of answering MCAT questions too!
A mass M moves from height = 0 to height = 5 m, and it does so straight vertically (as shown in the drawing with the pulleys). How you pull on it just doesn't matter. The force acting on it is that of gravity. Or alternatively you can just frame this in terms of potential energy: PE_1 = mgh = 4*0*10 = 0 and PE_2 = mgh = 4*5*10 = 200
Same as if I had a frictionless ramp of any angle. If I push something from the bottom to the ramp to a height of 5 meters, the work done is the same, regardless of the slope of the ramp.
Edit: I was silly, the rope itself would be pulled 5 meters. You could if you wanted to solve for the force first, and then the work done, but you would have to keep straight that there is now another actor not on the diagram doing the work and you're really calculating the work done by him or her. And for all that effort, you will find the two cos(60) terms canceling each other and your answer being identical. Identifying the efficient way of solving problems is a big part of answering MCAT questions too!