Question

Two blocks are connected by an ideal cord that passes over a frictionless pulley. If m₁ = 3.6 kg and m₂ = 9.2 kg, and block 2 is initially at rest 140 cm above the floor, how long does it take block 2 to reach the floor?

 

 

Answer 1

 

Let the acceleration of the two blocks is a.

The from the free body diagram we can have

T – m1g = m1a

And m2g – T = m2a

 

Adding the two equations we get m2g - m1g = m1a + m2a

Then (m2 – m1)g = (ma + m2)a

Or a = (9.2kg – 3.6kg)/(9.2kg + 3.6kg) (9.8 m/s2)

= 4.2875 m/s2

 

Now total distance moved by the block is 140m.

Now using the equation y = vayt + (1/2)ayt2

We get 140cm = (0m/s)(t) + (1/2)(4.2875 m/s2)(t2)

 

Or t = √(1.4m)(2)/(4.2875 m/s2)

= 0.81s

 

Therefore time taken by the block 2 to reach the ground is 0.81s.

Therefore time taken by the block 2 to reach the ground is 0.81s.