KCET · Physics · Laws of Motion
Block \(A\) of mass of \(2 \mathrm{~kg}\) is placed over block \(B\) of mass \(8 \mathrm{~kg}\). The combination is placed over a rough horizontal surface. Coefficient of friction between \(B\) and the floor is \(0.5\). Coefficient of friction between blocks \(A\) and \(B\) is \(0.4\). A horizontal force of \(10 \mathrm{~N}\) is applied on
block \(B\). The force of friction between blocks
\(A\) and \(B\) is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(100 \mathrm{~N}\)
- B \(40 \mathrm{~N}\)
- C \(50 \mathrm{~N}\)
- D Zero
Answer & Solution
Correct Answer
(D) Zero
Step-by-step Solution
Detailed explanation
Total mass of blocks \(A\) and \(B=2+8=10 \mathrm{~kg}\)
Friction between surface and combination of \(A\) and \(B\)
\(\begin{aligned}F &=\mu R \\&=0.5 \times 10 \times 10=50 \mathrm{~N}\end{aligned}\)
Here applied force on box \(B\) is \(10 \mathrm{~N}\) that is less than \(50 \mathrm{~N}\). So the system will be in rest because of this there is no friction between blocks \(A\) and \(B\).
Friction between surface and combination of \(A\) and \(B\)
\(\begin{aligned}F &=\mu R \\&=0.5 \times 10 \times 10=50 \mathrm{~N}\end{aligned}\)
Here applied force on box \(B\) is \(10 \mathrm{~N}\) that is less than \(50 \mathrm{~N}\). So the system will be in rest because of this there is no friction between blocks \(A\) and \(B\).
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