We are doing force and work right now and I'm completely lost. I don't even know what formulas I'm supposed to use or anything. Any help on getting any of these started or more would be greatly appreciated. Hopefully if I can get these first couple figured out, the rest will start to make sense. 1) A raindrop of mass m = 3.35 x 10^-5 kg falls vertically at constant speed under the influence of gravity and air resistance. After the drop has fallen 100m, what is the work done a) gravity and b) by air resistance. 2) A block of mass 2.50kg is pushed 2.20m along a frictionless horizontal table by a constant 16.0N force directed 25 degrees below the table. Determine the work done by a) the applied force b) the normal force exerted on the table, and c) the force of gravity. d) Determine the total work done on the block. 3) Batman, whose mass is 80.0kg is holding onto the free end of a 12.0m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough so that he can reach a ledge when the rope makes a 60 degree angle with the vertical. How much work was done against the force of gravity in this maneuver?

Work is force times distance in a given direction. Try this page: http://tutor4physics.com/workenergy.htm So to work out the first problem: a) gravity is pulling in the direction of travel of the raindrop. The force of gravity equals the mass of the object * the acceleration of gravity (F=ma) The work done by gravity would be the Force times the distance the drop travels, so: Work (in Joules) = Mass of object (in kg) * Acceleration of Gravity * distance traveled = (3.35 x 10^-5) * (9.8 m/s^2) * (100 m) b) Since the drop is at constant speed, acceleration = 0, so we can assume the force of the air resistance = the force of gravity. Thus, the amount of work of the air resistance would be exactly opposite that of gravity. (So just negative) The second problem is similar, but you have to figure out the horizontal and vertical components of the 16N force directed -25 degrees. The work done by the horizontal component would be Horizontal Force*Distance. The vertical component of the force is nullified by the table. There is no Work done by the force on the table or by gravity, because the block doesn't move in the vertical direction. Hope that helps some. Post back if you're still stuck.

I assume this is not first steps as that involves a few concepts not normally thrown in, I will not be giving direct answers either but hopefully it will be enough for you to follow. 1) has been taken care of already, air resistance = force due to gravity by the way. 2) a) Unusual wording, I run the risk of saying it was just 16N otherwise the force you need is that in the horizontal direction. cos25 = adjacent/hyp. -> h*cos25=a (if you are a bit sketchy on resolving forces and other such things I highly suggest working on it a lot (as in it is second nature a lot) for it will haunt you for a long time to come). b) This is the other side of the triangle, you can do h*sin25 = o or you can use pythag hyp^2= adj^2+ opp^2 , up to you which one although I suggest you have both on lock. c) gravity eqn: f=ma where a is acceleration due to gravity. d) They run the risk of bad wording again, assuming no deformation of the block it is just the standard work done = force x dist in the direction of the force (the direction part is very important and you should put it down when you define it). 3) Batman, whose mass is 80.0kg is holding onto the free end of a 12.0m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough so that he can reach a ledge when the rope makes a 60 degree angle with the vertical. How much work was done against the force of gravity in this maneuver? Another multi stage one. You need to know how far the rope has raised and then it is just potential energy gained (mass x gravitational field strength x height) This is another trigonometry question although perhaps a bit harder. My prefered way would be to imagine a triangle with 60 at the top with 2 sides being 12m drop a line down to make a right angled triangle to get 30 degrees and 12m hyp. sin30*hyp= length of base multiply by 2 to get length of other side of the original triangle make a new triangle with the hyp this new length Then by the fact that the angle above it 60 you can work out all the lengths you need. As an aside triangles based on angles 30, 60 and 45 degrees are special triangles that you can save some serious time on in exams (and they usually throw a few in to see) 45 degrees you have a box 1 by 1 andyou cut it along a diagonal. Pythag says that the diagonal is root 2 long meaning sine = opp over hyp which is 1 or root 2, likewise for cos. Tan is 1 over 1 (i.e. 1) 30 degrees and 60 degrees. Equilateral triangle 2 by 2 by 2, drop a line down from the apex to the base cutting it in half. This leaves you with a triangle 2 (hyp) by 1 by root 3 (2 squared minus 1 squared all square rooted). The small angle is 30, the big 60. 30 first. Opp= 1, adj= root 3 and hyp = 2 so sine= 1/2 cos=root3 over 2 and tan = 1 over root 3 60, basically the reverse of above/ I am bored and waiting to go shopping so you mgith as well have it. opp = root 3, adj= 1 and hyp = 2 sin= root3 over 2 cos = 1 over 2 and tan = root 3 over 1

Thanks for all the help guys. I haven't had a chance to look over your replies yet because I was finishing a vector project, but I'll get to it tonight when I get back from my next class.

Okay, so for #2a, cos25 = adjacent/hyp. -> h*cos25=a Cosine = adjacent / hypotenuse like in SOHCAHTOA But then h * cos25 = a Is that height times cosine25 = acceleration? What height? The only distance we have is the distance it traveled.

easy way to remember it: Silly Old Hitler Couldn't Advance His Troops Over Arabia Sin Opps Hypot Cosine Adjacent Hyp Tan Opps Adjacent

h=Hypotenuse, so I'm assuming it's 16*cos25. I could be wrong though, I haven't done mechanics in a while.

The physics of Batman... I'll give it an A for superhero relevance but an incomplete for terrible usage of the Dark Knight's grappling hook. Two answers come to mind for this problem: 1. The problem is trivial b/c there is an elephant (or other large obstacle in the way). Or... 2. It's another Force x Displacement problem. Going back to the question: How much work was done against the force of gravity in this maneuver? Against gravity tells you that the displacement needed is vertical and the amount of work needed can be calculated using Bat's mass x gravity x vertical displacement. Finding that vertical displacement... Maybe my crude ASCII representation can explain this: Initial state: | | | | | | Next state: (not to scale) (A) |\ | \ |__\ So B gets to the ledge when angle A is 60 deg. From the initial state (t=0), Batman is 12m away from the pivot point. At the next state(t=t1), his rope makes the 60 deg angle to the vertical, so the 12m is now the horizontal. Draw out the pic, label what you know and fill in the blanks. The vertical displacement should be there with some trig calculations and plug into the previously mentioned work equation. Hope this is right, The More You Know...

My bad, I speak of poorly worded questions and then I go and repeat the offence in my answer. When I spoke of trigonometry h and hypotenuse, a and adjacent and o and opposite were used interchangeably.

Hmm...I'm still confused for #2a. cos25 = adjacent/hyp. -> h*cos25=a Why are we using cos and where does hyp*cos25=a come from? and what's the hyp? don't we only know one side, the bottom (2.2m)? Sorry if I sound like an idiot, I just have such a hard time with math.

The triangle you are trying to figure out is to determine the horizontal component of the Force. The hypotenuse of the triangle is the 16N Force. The short leg is the vertical component, and the long leg is the horiz. component. So, the horizontal component of the force is 16N*cos25 - Answer (2a) The vertical component of the force is 16N*sin25 - Answer (2b) The work is then the product of the horiz Force and the distance traveled on the table. - Answer (2d)