Question

♡♡●In the system shown in the figure, the acceleration of the 1 kg mass and the tension in the string connecting between A and B is

Answer 1

From the free body diagram, we get

T+0.5a-0.5g=0-----------------(1)

μR +1a+T1 -T=0-----------------(2)

μR+1a-T1=0

μR +a= T1------------(3)

From equations i , ii and iii

μR+a=T-T1

T-T1=T1

T=2T1

Equation II becomes μR +a+T1-2T1=0

μR+a-T1=0

T1=μR+a

T1=0.2g +a

Equation I will become 

2T1+0.5a-0.5g =0

T1=0.5g -0.5a/2 

=0.25g-0.25a

From equation Iv and V

0.2g +a=0.25g -0.25 a

a=0.05/1.25 x10 

a=0.4 x 10

a=0.4 m/s2

Therefore acceleration of 1kg block is 0.4m/s2

b) Tension T1=0.2g +a+0.4=2.4 N

c) t=0.5g-0.5a

t=0.5x10-0.5x0l4

t=4.8N

Answer 2

Here is your answer in attachment❤