Question

Consider the situation shown in the figure (8-E2). The system is released from rest and the block of mass 1kg is found to have a speed 0⋅3 m/s after it has descended a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.
Figure

Answer 1

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Answer 2

The kinetic friction coefficient is 0.12

Explanation:

Step 1:

given values  

Now we have for 4 kg of block:- Initial velocity u = 0;  

distance traveled by 1 kg block = 1 m  

velocity = 0.3 m / s  

Now for a =?  

Step 2:

We are aware of the formula,

v² = u² + 2 as

v²- u² = 2 as

2 × a × 1 = 0.3² - 0

a = 0.09/2

a =0.045 m/s²

Step 3:

Tension T is the force on the surface, and friction F is the force

F = μmg

Now from Newton's rule of motion = 4 kg

T - F = m x 2 a

T - μmg= 4 x 2 x 0.045

T - 4μg = 0.36  

T = 0.36 + 4μg     ----------eqn (i)

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

ma = mg - T'

1 x 0.045 = 1 x g - 2T

0.045 = g - 2T

$T=\frac{g-0.045}{2}$     ----------eqn (ii)

Step 4:

The equation I & (ii) is equated and solved.

$\begin{equation}\begin{aligned}&0.36+4 \mu g=\frac{g-0.045}{2}\\&0.36+4 \times \mu \times 9.8=\frac{9.8-0.045}{2}\\&0.36+39.2 \mu=4.8775\\&\mu=\frac{4.8775-0.36}{39.2}\\&\mu=0.115\\&\mu=0.12\end{aligned}$

Therefore, the kinetic friction coefficient is 0.12