acceleration and tension

Question

 A 1,000 g block on a smooth table is connected to a 500-g piece of lead by a light cord that passes over a small pulley at the end of the table. a) What is the acceleration of the system? b) What is the tension in the cord?

Answer 1

a. The hanging weight is acted upon by two forces, the weight of the hanging object minus the tension force in the cord Fnet = 0.5 kg x 9.8 m/s² - T = 0.5 kg x a where a is the net downward acceleration.

The mass on the table (assumed frictionless) experiences a force equal to the tension force T = 1.00kg x a.

Putting T in the last equation into the first leads to 0.5 kg x a = 0.5 kg x 9.8 m/s² - 1.00 kg x a. Solving for

a = 3.27 m/s² which is the acceleration of the system.

The tension in the cord using the first equation above is 0.5 kg x 3.27 m/s² = 0.5 kg x 9.8 m/s² - T or T = 3.27 N

or from the second equation above T = 1.00 kg x 3.27 m/s² = 3.27 N

Answer 2

Suppose T is tension in string, and m1 is mass of 1000g block and m2 is mass of 500g block

Then

net force on mass 1 = T = m1 a

net force on mass 2 = (m2 g - T) = m2 a

substituting for T

(m2 g - m1a) = m2a

So a = m2 g / (m1 + m2)

m1 = 1 kg

m2 = 0.5 kg

so a = 3.27 m/s/s with g = 9.81

T = m1 a so T = 3.27 N