Question

Two blocks with masses m1 and m2 of 10kg and 20kg respectively are placed as in fig u=0.2 between all surface,then tension in string and acceleration of m2 block at this moment will be​

Answer 1

Given :

The mass of block 1 = $m_1$ = 10 kg

The mass of block 2 = $m_2$ = 20 kg

The value of $\mu$  between surface = 0.2

To Find :

The tension in string

The acceleration of block 2 at this moment

Solution :

Free body diagram of both the blocks are given in figure

For block $m_1$ , Let T force is applied on vertical position

Total force at equilibrium

T sin 30°= mass × g

i.e  T × 0.5 = 10 × 9.8 m/s²

∴  String Tension = 196 Newton

 Again

 Horizontal component of Tension of the string applying force on block 2

∵ For block 2 of mass $m_2$ , Let N is normal reaction force is applied on horizontal position

Total force at equilibrium

N  = T cos 30°

i.e  N = 196 × $\dfrac{\sqrt{3} }{2}$

∴   N = 169.7 Newton

So, Tension applied on string = N = 169.7 Newton

Again

For block 2 , Applying vertical force direction, we get

20 × g - 2 × μ × N  = 20 × a

Substituting the value we get ,

20 × 9.8 m/s² - 2 × 0.2 × 169.7 Newton  = 20 × a

Or,  196 - 67.88 = 20 a

Or, 20 a = 128.12

∴         a = $\dfrac{128.12}{20}$

i.e acceleration = a = 6.40 m/s²

Hence, The tension force on string is 169.7 Newton

And The acceleration of block 2 at this moment will be​ 6.40 m/s² . Answer