Section Student Name Report on Laboratory Experiment \"Newton\'s Second Law: Atw
ID: 1881058 • Letter: S
Question
Section Student Name Report on Laboratory Experiment "Newton's Second Law: Atwood's Machine" We can leave all masses in grams (3) since all calculations involve ratios of masses. DATA TABLE Trial 1 Trial 2 Trial 3 Trial 4 Mass of pulley Mg Mass of the first weight holder mheweight holder 1 (8) Mass of the weight, placed in the weight holder 1 Ann (g) Total ascending mass mi (g) Note Mass of the second weight holder mpeight holder 2 Mass of the weight, placed in the 73 401.5 gs.sz. Ilo2.5T 123.5% S3 715 weight holder 2 4m2 (g) Total descending mass m2 Height of fall h Measurement 1 Measurement 2 Measurement 3 Measurement 4 Measurement 5 Average time of fall t (s) Theoretical acceleration a from (1) (m/s) Measured acceleration a from (2) m/s) % DifferenceExplanation / Answer
BD equation of m1 , T-m1g =m1a
where T is tension in the string , a is acceleration of m1
FBD equation of m2 , m2g-T =m2a
adding both equations we get
m2g-m1g =m1a+m2a
(m2-m1)g=(m1+m2)a
acceleration a = (m2-m1)g/(m1+m2)
for trial 1 , given m2 =73.71 g m1 =63.58 g
a = (73.71-63.58 )9.8 /(73.71+63.58)
a=0.723 m/s^2
For trial 2, m2=93.71 m1 =83.58
a =(93.71-83.58)9.8/(93.71+83.58)=0.56 m/s^2
For trial 3 , m2 =113.71 m1 =103.58
a=(113.71-103.58)9.8/(113.71+103.58)=0.457 m/s^2
for traial 4 , m2 =133.71 m1 =123.58
a = (133.7-123.58)9.8/(133.71+123.58) =0.386 m/s^2
distance travelled by mass ,H =0.5 at^2
given H=0.89 m
For trial 1 , 0.89= 0.5*0.723 *t^2
t^2 =0.89/(0.723*0.5)
t^2=2.462
t= sqrt (2.462)
time t = 1.569 seconds
for trial 2. a=0.56m/s^2 , H=0.5at^2
t=sqrt(2H/a) =sqrt((2*0.89)/0.56) =3.56 sec
For trial 3 a =0.457
t =sqrt(H/a)= sqrt(2*0.89/0.457) =6.96 sec
For trial 4 a= 0.386
t=Sqrt(2H/a)= sqrt(2*0.89/0.386) =18.01 seconds
now i order to find theoritical acceleration from m1 measure the time for fall and find the accelration using H=0.5at^2 and calcute difference between theoritical and measured values of acceleration .
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