- #1

kieyard

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**THIS THREAD WAS INCORRECTLY PLACED IN THE CLASSICAL PHYSICS FORUM, SO THERE IS NO TEMPLATE**

Hi, everyone.

my teacher stuck me with these questions and i am need of a little help answering question 2.

here's my answers to question 1 first.

1a) a=0, as system would be in equilibrium when F=0

b) F=5mgN, as for a=5m/s the acceleration of B has to equal 2.5m/s and as its in equilibrium F has just got to equal m*a which is 2.5*2mg=5mgN

c) F=mga

i think my answers are right but if I am making a simple mistake let me know.

now my attempts at 2.

2a) first, B has to x/2 above the ground for the full system to work so original potential energy is 2mgx/2=mgx assuming A starts on the ground. now its the following kenetic energy bits that are confusing me because there will be two different V's one for A and one for B. V

_{A}= 2V

_{B}so assuming no energy gets lost to heat via friction whatever all of the gravitational energy plus energy input from F has to turn into kinetic energy so I am thinking.

mgx + Fx = 1/2mV

_{A}

^{2}+ mV

_{B}

^{2}

which kind of works if your given enough information. but part B doesn't i don't think.

2b) substituting values into my answer to 2a

98 + 196 = 1/2[ V

_{A}

^{2}+ 2V

_{B}

^{2}]

so V

_{A}

^{2}+ 2V

_{B}

^{2}= 588N

and V

_{A}= 2V

_{B}

so with further manipulation i get V

_{A}= 19.8m/s

but using logic from question 1. if F=19.6 and B has a mass of 2kg, the acceleration of B = 9.8m/s

^{2}therefore acceleration of A = 19.6m/s

^{2}and then using SUVAT.

s=10m u=0 v=? a=19.6 m/s

^{2}

v

^{2}= u

^{2}+2as

v = 23.2m/s

which is different. iv tried multiple times and i can't figure out what I am doing wrong. I am pretty sure its the WEP part not the SUVAT part as I am better at that but i do not know.

please help.

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