Question

Consider the situation shown in figure. The system is released from rest and the block of mass 1.0 kg
is found to have a speed 0.3 m/s after it has descended through a distance of Im. Find the coefficient
of kinetic friction between the block and table​

Answer 1

Answer:

Now for 4 kg of block we have :-

Initial velocity u = 0 , 

Distance travelled by 1 kg block = 1 m

velocity = 0.3 m/s

Now for finding acceleration a =?

We know the formula,

v² = u² + 2as

v²- u² = 2as

2 × a × 1 = 0.3² - 0

a = 0.09/ 2

a =0.045 m/s²

Here force on the block are Tension T , and force on the Friction F 

F = μmg

Now from Newton law of motion :- For 4 kg

T - F = m x 2a

T - μmg= 4 x 2 x 0.045

T - 4μg = 0.36 

T = 0.36 + 4μg ----------→ (i)

Newton law of motion for 1 kg block :- For tension T' = 2T

ma = mg - T'

1 x 0.045 = 1 x g - 2T

0.045 = g - 2T 

T =(g - 0.045)/2 ----------→ (ii)

On equating and solving equation (i) & (ii) we get,

0.36 + 4μg = (g- 0.045 )/2

0.36 + 4 x μ x 9.8 =(9.8 -0.045) /2

0.36 +39.2 μ =  4.8775

μ =(4.8775 - 0.36) / 39.2

μ = 0.115 or 

μ = 0.12

Hence the coefficient of Kinetic friction is 0.12

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