Two masses are connected by a string. One mass m2 rests on a frictionless table,

Q: Two masses are connected by a string. One mass m2 rests on a frictionless table,

a) Let a is the acceleration of blocks. LEt T is the tension in the string. net force acting on m1, Fnet1 = m1*g - T m1*a = m1*g - T T = m1*g - m1*a ---(1) Net force acting on m2, Fnet2 = T m2*a = T ----(2) from equations `1 and 2 m2*a = m1*g - m1*a a*(m1+m2) = m1*g a = m1*g/(m1+m2) = 50*9.8/(50+100) = 3.27 m/s^2 b) a = m1*g/(m1+m2) = 65*9.8/(65+85) = 4.25 m/s^2

ID: 1336923 • Letter: T

Question

Two masses are connected by a string. One mass m2 rests on a frictionless table, and the other mass m1 is suspended from the edge of the table. The string passes over a frictionless pulley at the edge of the table (see figure 6.2 in the lab manual).

a) If m1 = 50 g and m2 = 100 g, what is the acceleration of the masses immediately after they are released from rest?

b) If15gistransferredfromm2 tom1,suchthatm1 =65gandm2 =85g, what is the acceleration of the masses immediately after they are released from rest?

Explanation / Answer

a)
Let a is the acceleration of blocks.

LEt T is the tension in the string.

net force acting on m1, Fnet1 = m1*g - T

m1*a = m1*g - T

T = m1*g - m1*a ---(1)

Net force acting on m2, Fnet2 = T

m2*a = T ----(2)

from equations `1 and 2

m2*a = m1*g - m1*a

a*(m1+m2) = m1*g

a = m1*g/(m1+m2)

= 50*9.8/(50+100)

= 3.27 m/s^2

b) a = m1*g/(m1+m2)

= 65*9.8/(65+85)

= 4.25 m/s^2

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Two masses are connected by a string which passes over an ideal pulley as shown

Two masses are connected by a string. One mass m2 rests on a frictionless table,

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Two masses are connected by a string. One mass m2 rests on a frictionless table,

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