Question

shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood’s machine. One block has mass m1 = 1.30 kg; the other has mass m2 = 2.80 kg. What are (a) the magnitude of the blocks’ acceleration and (b) the tension in the cord?

Answer 1

Answer:

a = 3.585 m/s^2

T = 17.4 N

Explanation:

Given:

- The mass of block 1 m1 = 1.30 kg

- The mass of block 2 m2 = 2.8 kg

- The acceleration of block-block system = a

- The tension in the cord = T

Find:

(a) the magnitude of the blocks’ acceleration

(b) the tension in the cord?

Solution:

- In our coordinate systems, the acceleration of the block 1 equals the acceleration of the block 2. So we can express them as a common acceleration a. The equations of motion are :

                                     m1*a = T - m1*g

                                     m2*a = m2*g - T

- Adding the two equations together, we get the equation of motion for the whole system.

                                  (m1+m2)*a = m2*g-m1*g

- Solving this equation for the acceleration, we have

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

                                  a = (2.8-1.3)×9.8/(1.3+2.8)

                                  a = 3.585 m/s^2

- Plugging the result into the equation of motion for block 1, we have

                                  T = m1*(a+g)

                                  T= 1.3×(3.585+9.8)

                                  T= 17.4 N

Answer 2

(A) When the magnitude of the blocks’ acceleration is = $3.585 m/s\wedge 2$

(B) The tension in the cord T is = 17.4 N

What is an Acceleration?

Given as per question are:

The mass of block 1 m1 is = 1.30 kg

The mass of block 2 m2 is = 2.8 kg

The acceleration of the block-block system is = a

The tension in the cord is = T

(A) In our coordinate systems, When the acceleration of block 1 equals the acceleration of block 2. Then, we can express them as a common acceleration a. Now, The equations of motion are :

After that, m1*a = T - m1*g

Then, m2*a = m2*g - T

After that Adding the two equations together, then we get the equation of motion for the whole system.

Using equation of motion is = (m1+m2)*a = m2*g-m1*g

Then we solving this equation for the acceleration, we have

Now, a = (m2-m1)*g/(m1+m2)

After that, a = (2.8-1.3)×9.8/(1.3+2.8)

Thus, a is = $3.585 m/s\wedge 2$

(B) Plugging the result into the equation of motion for block 1, we have

After that, T = m1*(a+g)

Then, T= 1.3×(3.585+9.8)

Therefore, T= 17.4 N

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