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do you know how to define weight using mass?

hmm... good one.

w = mg

velocity . time graph

|dw:1349878299626:dw|

how come constant velocty?

\[a = \frac Fm\]
\[a = \frac {\frac59mg}m\]
\[a = \frac59g\]

\[v=\frac{5}{9}t+v_o\]
can we assume an initial velocity of 0?

yes lets see if the answer comes

i dont think, the answer is 7/3 mg

i havent gotten to the end yet ... just wondering if my thought process made sense so far

prolly a bad graph :/

hmm

i got the answer thanks :)

u wanna see my working

|dw:1349879825375:dw|

dude why are you posting questions you don't need help on?

let's see your second part of the soln. I hope your first part agrees with mine?

wht do umean? which part do u need to agree

...

did you get the same results as I did for the first part of the descent, where a= -4g/9?

yes :)

i substitues and got friction as 7/3 mg

wut.

you got the force from friction = 7mg/3 ??

yes i did

ok. that means a net force upwards. since the force down is mg

pretty funny.

yeah..could u help me with the second part i posted right now :)

you make no sense.

what acceleration are you using for the first part?

4/9 g

and when the friction increases it more thn the weight, acceleration becomes - 4/3g

by negative I guess you mean upwards.

deceleration

it's just a matter of plugging in 'h'

no its correct...

I mean, you're basically just assuming the whole trip takes 1 second...

yup

so you plotted
v =( (3/9) g)*t

yes...

so second part.

yes how do i do tht

oh okay il try

btw which equation did u use

sorry about the confusion, I totally misinterpreted that problem I guess.

I used
d= Vi*t + 1/2*a*t^2

d1= 1/2* 4g/9*(3t/4)^2
d2 = (3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2
d1 + d2 = 60

ok

you get it? I got t=8 (using g=10)
and V = 24/9*g