Question

Consider the situation shown in 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.

Answer 1

Given,

$m_1=4kg,m_2=1kg,$

$v_2=0.3m/sec$

$v_1=2\times \left(0.3\right)=0.6m/sec$

$x_1=2x_{2}m$ in this system

h = 1 m = height descent travelled by 1 kg block = $x_2$

$x_1=2\times 1=2m$ = distance travelled by 4 kg block

Taking $u=0$ and

Applying, change in K.E. = Work done (for the system)

$\left(\frac{1}{2}\right)m_1{v}_1^2+\left(\frac{1}{2}\right)m_2{v}_2^2=(-\mu R)x_1+m_{2}gh$

[R = 4g = 40 N]

$\Rightarrow \frac{1}{2}\times 4\times \left(0.36\right)+\frac{1}{2}\times 1\times \left(0.09\right)=-\mu \times 40\times 2+1\times 10\times 1$

$\Rightarrow 0.72+0.045=-80\mu +10$

$\Rightarrow \mu =\frac{8.83}{80}=0.11$