Question

A 4 kg block sits at rest on a frictionless table. It is connect by a light string over an ideal pulley to a 2 kg block hanging off the edge of the table. If the 2 kg block is released, how long will it take to fall 56 cm????

Answer 1

for hanging mass of 2 kg ,

tension (T) in the string = m(g-a)

= (2)(g-a)

= 2g - 2a ..(1)

for stationary block of mass 4 kg

tension (T) in the string = m1a

= 4a ..(2)

equating equations 1 and 2

2g - 2a = 4a

2g = 6a

g/3 = a

9.8/3 = a

a = 3.27 m/s^2

therefore, acceleration of both the masses is equal to 3.27 m/s^2

now, d = 56 cm = 0.56 m

using equation of motion , d = ut + (1/2) at^2 ; where u is the initial velocity = 0 m/s

d = (1/2) at^2

0.56 = (1/2)(3.27)t^2

t = 0.585 seconds.

{ time taken by hanging mass to fall a distance equal to 56 cm}

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