solve this quesstn .... 34 no. plzzz
Answers
pls refer the diagram first for better understanding
First we need to find maximum frictional force between the blocks
Between 10 kg and 20 kg block
f 1 max = u N
f 1 max = 0 .5 ×10×10
f 1 max = 50 N
Between 20 kg and 30 kg block
f 2 max = u N
f 2 max = 0 .25 ×(10+20)×10
f 2 max = 0 .25×30×10
f 2 max = 75 N
Forces acting on 10 kg block
100 - f 1 = m a
100 - f 1 = 10 a ...(1)
f 1 = ( 20 + 30) a
f 1 = 50 a
a = f 1 / 50 ....(2)
Substituting the value of a in eq 1
100 - f 1 = 10 f 1 / 50
100 = 10 f 1 + 50 f 1 / 50
5000= 60 f 1
f 1 = 83.33 N
f 1 > f 1 max
( 10 kg block will undergo slipping)
Forces acting on 20 kg block
f 1 - f 2 = 20 a
50 - f 2 = 20 a ....(3)
f 2 = m a = 30 a
a = f 2 / 30 ...(4)
Substituting the value of a in eq 3
50 - f 2 = 20 f 2 /30
50 ×30= 20 f 2 + 30 f 2
1500 = 50 f 2
f 2 = 1500 /50
f 2 = 30 N
f 2 < f 2 max
( hence no slipping will take place . 20 kg block and 30 kg block will move together)
In order to find the acceleration of the 20 kg block
Substituting the value of f 2 in eq 3
50 - f 2 = 20 a
50 - 30 a = 20 a
50 = 50 a
a = 1 m / s ²
