Question

Two identical blocks are attached to the same massless rope, which is strung around two massless, frictionless pulleys. A massless scale is connected to the same rope and measures the tension in the rope. The two identical blocks are released from the rest. The experiment is then repeated with two new blocks, with masses m1 and m2. When they are released from rest, the system remains at rest, and the scale measures the same tension as in the previous experiment. Find m1 and m2 in terms of m.

Answer 1

Answer:

The value of mass 1, m1= 6/5m

The value of mass 2, m2= 3/5m

Explanation:

case 1:

here tension and the acceleration will be:

for m1;

• mg-T=ma
• 2mg - 2T = 2ma  .....1.

for m2:

• 2T-mg = ma/2 ..... 2.

adding the both equations,

2mg - 2T + 2T-mg = 2ma + ma/2

a = 2/5 g

putting the value of a into the equation 1.

mg - T = m* (2/5)g

T = 3/5 ( mg )

now

case 2:

The two identical blocks are released from the rest, the tension remains the same as the case 1.

so,

for m1:

• 2T-m2g=0

for m2:

• 2m2g - 2T =0

adding both equations we get,

2T-m2g + 2m2g - 2T = 0

m2 = m1 / 2

T = m1*g / 2

here we know that

T (case1) = T (case2)

3/5 ( mg ) = m1*g / 2

m1 = 6/5 m

hence

m2 = 3/5 m

learn more about tension here:

brainly.com/question/23590078

#SPJ4