TS EAMCET · Physics · Laws of Motion
A block of mass \(m=2 \mathrm{~kg}\) is initially at rest on a horizontal surface. A horizontal force \(\mathbf{F}_1=(6 \mathrm{~N}) \hat{\mathbf{i}}\) and a vertical force \(\mathbf{F}_2=(10 \mathrm{~N}) \hat{\mathbf{j}}\) are then applied to the block. The coefficients of static friction and kinetic friction for the block and the surfaces are 0.4 and 0.25 , respectively. The magnitude of the frictional force acting on the block is (assume, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )

- A \(2.5 \mathrm{~N}\)
- B \(4.0 \mathrm{~N}\)
- C \(3.3 \mathrm{~N}\)
- D \(3.0 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(2.5 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
In addition to the forces shown in figure, there will be another upcoming normal force \(\mathbf{F}_N\) exerted by the floor on the block. Using Newton's second law in horizontal and vertical directions, we get In horizontal direction, \(F_1-f_s=m a\) \(\ldots(\mathrm{i})\) and…
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