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elixir [45]
3 years ago
10

If one of the masses of the Atwood's machine below is 2.9 kg, what should be the other mass so that the displacement of either m

ass during the first second following release is 0.28 m? Assume a massless, frictionless pulley and a massless string. (There are two possible answers for m2. Enter your answers from smallest to largest.)
Physics
1 answer:
Afina-wow [57]3 years ago
8 0

Answer:

2.59 Kg, 3.25 Kg

Explanation:

Acceleration can be found using equation

s=0.5at^{2} where s is the release distance, a is acceleration and t is time

Making a the subject of the formula

a=\frac {2S}{t^{2}}

Substituting 0.28 for s and time for 1 second

a=\frac {2*0.28m}{(1s)^{2}=0.56 m/s^{2}

Acceleration formula for the Atwood machine is given by

a=\frac {g(m1-m2)}{m1+m2}  where m1 and m2 are first and second masses respectively

Two situations are possible

When m1>m2

Assuming m1 is 3.7kg which is heavier than m2

Substituting a for 0.56 m/s^{2}  and m1 as 2.9Kg and taking acceleration due to gravity as 9.81 m/s^{2}  

0.56 m/s^{2}=\frac {9.81 m/s^{2}*(2.9Kg-m2)}{2.9Kg+m2}  

(0.56 m/s^{2})*(2.9Kg+m2)=9.81(2.9Kg-m2)  

(0.56 m/s^{2}+9.81)*m2=(9.81*2.9)-(2.9*0.56)  

10.37m2=28.449-1.624

10.37m2=26.825

m2=\frac {26.825}{10.37}=2.5867888138  

m2=2.59 Kg

<u>When m1<m2</u>

a=\frac {g(m2-m1)}{m1+m2}  

Then m1=2.9Kg hence

0.56 m/s^{2}=\frac {9.81(m2-2.9Kg)}{2.9Kg+m2}

(0.56 m/s^{2})*(2.9Kg+m2)=9.81 m/s^{2}*(m2-2.9Kg)

(0.56 m/s^{2}-9.81 m/s^{2})*m2=(-2.9 Kg*9.81 m/s^{2})-(2.9 Kg*0.56 m/s^{2})  

-9.25m2=-28.449-1.624=-30.073

9.25m2=30.073

m2=\frac {30.073}{9.25}=3.2511351351

m2=3.25 Kg

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