Recent content by Underdog_85

I got 4300J. It does say 10700J in book. It is mistake.

I got 10.25m for the vertical.. I popped that in the W=Fs and W=FsCostheda equations but I still don't get 10 700J.

I don't know how to get the vertical.. There is no example. Would I use trig to find the vertical? hypotaneuse being 25, inside angle being 25, and figure out the opposite side(vertical)?

Question 2 W=Fs(costheda) W=(200N)(25.0m)(cos25.0degrees) W=4531 N This is incorrect though.. I don't know.. This is just frustrating me now. They don't give examples for this in the book.

Ok I am getting two different answers and it is just confusing me. This is what I did for 1 a) W=Fs W=(215N)(4.20m) W=903Nm or J 1 b) W=Fs W=(-215N)(4.20m) W=-903Nm or J These are both the final answers in the book, however I am still confused about what we do with the horizontal 6.80m.

I wouldn't agree with you. If the displacement is 0 than the work would be 0, however there is a displacement of 4.20m, plus the final answer is also 903 Joules or Newton Metres in the back of the text. So the answer cannot be 0.

Two Work problems I can't figure out. I don't really know where to start? Question1) A delivery person carries a 215N box up stairs 4.20m vertically and 6.80m horizontally. a)How much work does the delivery person do?(Answer: 903 J) b)How much work does the delivery person do in carrying the box...

1. A ball weighing 520g is thrown straight up at 6.0m/s a) What is the initial momentum of the ball in N.s 2. p=mv =520g(1kg/1000g)(6.0m/s) =3.12 =3.1 N.s I don't see how this is correct because it takes kg.m.s^2 to create a Newton. There is none here. Then we go on to...

Q1: Assuming friction force to be negligibly small, what will be the acceleration in m/s^2 of a 45kg skater when propelled by a force of 250N? My attempt Data: F-250N m-45kg a-m/s^2 Formula: F=ma a=F/m a=250N/45kg a=5.555 a=5.6m/s^2 Q2: a 0.5kg mass on a horizontal frictional free...