Question

Consider the situation shown in the figure (8-E2). 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 1 m. Find the coefficient of Kinetic Friction between the block and the table.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question :-

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


Hope it Helps.

Answer 2

Answer:

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