Question

Two blocks A and B, each of mass 10kg, are
sliding on an incline. They are connected by
a string. The coefficients of kinetic friction
between block A and incline is M = 0.and
that between block B and incline is M2 = 0.4.
The tension in string will be (take g = 10
m/s2)
TAZ
B
37°
Zero
8N
4N​

Answer 1

Given that,

Mass of block A = 10 kg

Mass of block B = 10 kg

Angle = 37°

Coefficient of kinetic friction between A and incline = 0.2

Coefficient of kinetic friction between B and incline = 0.4

We need to calculate the acceleration

For mass A

Using balance equation

$m_{A}a=m_{A}g\sin\theta-T-f$

$m_{A}a=m_{A}g\sin\theta-T-\mu(m_{A}g\cos\theta)$

Put the value in the equation

$10a=10g\sin37-T-0.2\times10\times10\cos37$.....(I)

For mass B

Using balance equation

$m_{B}a=m_{B}g\sin\theta-T-f$

$m_{B}a=m_{B}g\sin\theta+T-\mu(m_{B}g\cos\theta)$

Put the value in the equation

$10a=10g\sin37+T-0.4\times10\times10\cos37$.....(II)

Now, from equation (I) and (II)

$20a=120-47.94$

$a=\dfrac{120-47.94}{20}$

$a=3.6\ m/s^2$

Put the value in the equation (I)

$10\times3.6=10\times10\sin37-T-0.2\times10\times10\cos37$

$-T=36-79.9+15.98$

$T=27.92\ N$

Hence, The tension in string will be 27.92 N.