Question

to phi @Mathematics

Answer 1

pls post the ansers as a doc

Answer 2

i will be online tmmrw this time can u giv answers thn??

Answer 3

solution to Q5:
the net force acting on the body will b the vector sum of all the forces acting on the body and this will provide the necessary acceleration for the body
F=ma
a=F/m
\[F=\sqrt{6^{2}+8^{2}}\]
i.e. F=10N
this implies a=10/2
\[a=5ms ^{-2}\] along
\[\tan^{-1} 3/4\]

Answer 4

q1 to q15, q1 at the end.

Answer 5

thanks a lot phi!!!!

Answer 6

u are the BEST

Answer 7

Some comments on my post: sometimes I did not label the units. This is a bad habit. Sometimes I mis-label acceleration m/s, when it should be m/s^2
In some of the problems I found force = g, These should be written as 9.8 N, because g is generally understood to be an acceleration.

The last line of answer 1 has an extra dt in it. It should read
dp/dt= mass*dv/dt s= 1575 kg/s * 1000 m/36 s = 43750 kg-m/s^2= 43750 N
where mass = dt *1575 kg

Answer 8

:) the effort u hav taken. i can neglect the mistake lol.. by the by u drm MIT??

Answer 9

phi??

Answer 10

If you want to solve these problems quickly, note that most can be categorized into 2 types.

Q 2,3,5,9,10,11,12,13
Find the net force on the system (all the masses together)
Use the system's net force to find the system's acceleration: a_sys= F/M
For an individual block, find its net force: F_block = m_block*a_sys
Use the equation F_left+F_right= F_block to find e.g. tension

NOTE: the pulleys in these problems are a distraction. change the problem to a simple horizontal line of blocks tied together by a rope.

Use sign conventions: + is to the right, - is to the left. (Tension is both)

Force parallel to a slope is m g sin(theta) where theta is the slope
Force normal to a slope is m g cos(theta)

Q4,6,7,10,14,15
Friction problems
Find the force Fn normal to the surface
Find the force Fp parallel to the surface (sometimes)
Friction= mu*Fn (this is a max force. friction acts against a force f (= to f), up to this max)
use F_net+F_right+F_left= 0
where F_net is the net force on a block: F_net= mass_block*a_block

Q6 was more complicated. Q1 did not fit into either of the above categories.

Answer 11

:) wel one thing pal the tension in 2nd qn was not 33

Answer 12

what was it?

Answer 13

1300/3

Answer 14

phi??

Answer 15

I'm thinking about that. Can you show the work?

Answer 16

k come here http://www.twiddla.com/671181

Answer 17

Fixed up Q2. I noticed that was the first one I did. I got better as I went along.

Answer 18

Here's the energy/work problem

Answer 19

is there crction on any othr question than 2??

Answer 20

Don't you have the answers?

Answer 21

wel not of every qns

Answer 22

I didn't find any, but you never know.

Answer 23

hey check out that ratio qn?

Answer 24

u got 5 in that i think its 1:1

Answer 25

Q7. It asks for the ratio of the powers needed in the above cases.
I obviously found the ratio of the forces.

Answer 26

I'm not sure what they meant by powers.

Answer 27

P= Force * distance ? rings a bell.

Answer 28

or Power = force*distance/time
??

Answer 29

2nd one

Answer 30

f*d/t

Answer 31

The question states "moves with the same speed" so d/t is the same for both.
so it comes to the ratio of the forces. If so, that is not 1/1
Any other ideas?

Answer 32

it is 1:1 i suppose

Answer 33

It's obvious the force need to move up hill is greater than to move down hill.

Answer 34

next qn???

Answer 35

What is it?

Answer 36

Can you repost as a separate question. This one is way too long

Answer 37

k