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?

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

b) 10 minutes

girls stopped for 10 minutes.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach