how to find tension between two blocks on a pulley
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Tension is constant across the string. Hence,
T - mg = ma (1)
Mg - T = Ma (2), where M and m denote large and small mass respectively
Add equations (1) and (2):
g(M - m) = a(m + M)
a = g(M - m)/(m + M)
Also, T = m(g+a) = m*g(1+(M-m)/(m+M).
Now plug the values!
T = 1.00 kg * 10m/s^2 * (1 + (2.00 kg - 1.00 kg)/(2.00 kg + 1.00 kg))
= 10 kg m/s^2 * (4/3)
= 13.3 N (to 3 significant figures)
Note: I assumed that the values were to three significant figures. If the question was worded as written above the answer would just be 13N.
HOPE IT WILL HELP U:)
T - mg = ma (1)
Mg - T = Ma (2), where M and m denote large and small mass respectively
Add equations (1) and (2):
g(M - m) = a(m + M)
a = g(M - m)/(m + M)
Also, T = m(g+a) = m*g(1+(M-m)/(m+M).
Now plug the values!
T = 1.00 kg * 10m/s^2 * (1 + (2.00 kg - 1.00 kg)/(2.00 kg + 1.00 kg))
= 10 kg m/s^2 * (4/3)
= 13.3 N (to 3 significant figures)
Note: I assumed that the values were to three significant figures. If the question was worded as written above the answer would just be 13N.
HOPE IT WILL HELP U:)
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