•51 GO Figure 5-47 shows two blocks connected by a cord (of negligible mass) that passes over a fric- tionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. One block has mass m¡ = 1.30 kg; the other has mass m2 2.80 kg. What are (a) the magnitude of the blocks' ac- celeration and (b) the tension in the cord? m1

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Chapter1: Units, Trigonometry. And Vectors
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51
•51 GO Figure 5-47 shows two blocks connected by
a cord (of negligible mass) that passes over a fric-
tionless pulley (also of negligible mass). The
arrangement is known as Atwood's machine. One
block has mass m¡ = 1.30 kg; the other has mass m2
2.80 kg. What are (a) the magnitude of the blocks’ ac-
celeration and (b) the tension in the cord?
m1
M2
Transcribed Image Text:•51 GO Figure 5-47 shows two blocks connected by a cord (of negligible mass) that passes over a fric- tionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. One block has mass m¡ = 1.30 kg; the other has mass m2 2.80 kg. What are (a) the magnitude of the blocks’ ac- celeration and (b) the tension in the cord? m1 M2
Expert Solution
Step 1

Given:

mass, m1 = 1.30 kg

mass, m2 = 2.80 kg

 

Solution:

                                                Physics homework question answer, step 1, image 1

From the free-body diagram of the system, we get:

            T - m1g = m1a      -1m2g-T = m2a        -2

 

From equation 1 and 2 , we get:

    T = m1a + m1g              -3T = m2g- m2a               -4

Equating equation 3 and 4 , we have:

m1a + m1g = m2g- m2a                 m1a+m2a   =m2g- m1g a m1+m2=gm2-m1a= g m2-m1m1+m2

Substituting the given values in above equation, we get:

a= 9.8 m/s2 2.80 kg-1.30 kg2.80 kg+1.30 kg=0.366 × 9.8 m/s2 =3.59 m/s2

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