answer in m/s2

Question

A block of mass 4.29 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 3.61 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the the pulley to have no friction and no rotational inertia. The acceleration of gravity is 9.8 m/s2 Calculate the acceleration of the first block. Answer in units of m/s .

Answer 1

Block mass is M = 4.29 kg, hanging mass m = 3.61 kg.

Since the surface of the table is smooth, the system (m+M) will move with acceleration a.

According to Newton's second law

Σ F= (mass)(acceleration)

T- T + (mg) = (m+M)a

Which means:

The tension of the rope  in the system in the x direction cancels with the tension vertically:

 

T-T +3.61 g = ( 3.61 + 4.29) a 

(3.61) (9.8) = 7.9 a

 

a= 35.38/7.9 =  

 

a = 4.48 m/s^{2 in the direction of motion (downwards).}