Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
[tex]T-f-T'=m_Ba[/tex]
Where [tex]m_B[/tex]=Mass of block B
Substitute the values
[tex]50-10-T'=m_B\times 0=0[/tex]
[tex]T'==40 N[/tex]
For block A
[tex]T'-f'=m_Aa[/tex]
Where [tex]m_A=[/tex]Mass of block A
Substitute the values
[tex]40-f'=m_A\times 0=0[/tex]
[tex]f'=40 N[/tex]
Hence, the friction force acting on block A=40 N