Question

5

The system shown is released from rest and moves 80 cm. What is the final speed, if m_1 = 3kg, m_2 = 2kg, frictional surface with mu =0.3, and pulley as solid disk with M = 1.5kg and R = 0.3 m?

Answer 1

Answer 2

SOLUTION :

Consider pulley be frictionless, so tension before pulley and 

after pulley remains same.

Aș per forces acing on block M1 :

M1 g - T = M1 a 

=> 3 *9.8 - T = 3 a

=> 29.4 - T = 3 a ……..……. (1)

Aș per forces acing on block M2 :

T - µk * M2 g = M2 a 

=> T - 0.3*2*9.8 = 2 a 

=> T - 5.88 = 2 a     ………. (2)

Add (1) and (2) :

=> 23.52 = 5 a 

=> a = 23.52/5 = 4.704 m/s^2

Let final velocity after 80cm movement be v m/s.

So,

v = 0 + a t = 4.704 t       m/s

Average velocity during interval t seconds 

= (0 + 4.704 t) / 2 

= 2.352 t       m/s

Distance moved in t sec. = avg. velocity * t

=> 80/100 = 2.352 t * t 

=> t = sqrt(0.8/2.352) 

=> t = 0.5832 sec.

So,

final velocity after 80 cm movement 

= v

= a t 

= 4.704 * 0.5832

= 2.74 m/s (ANSWER).