A block of mass 2 kg rests on a horizontal surface. If a horizontal force of 5 N is applied on the block the frictional force on it is (μk=0.4, μs=0.5)

Question

A block of mass 2 kg rests on a horizontal surface. If a horizontal force of 5 N is applied on the block the frictional force on it is (μk=0.4, μs=0.5) (A) 5 N (B) 10 N (C) 8 N (D) zero

Tardigrade Admin · Accepted Answer

Minimum force required to move the block =μs m g=0.5 × 2 × 9.8=9.8 N Since the force applied is only 5 N, the block fails to move. ∴ Frictional force = applied force =5 N

Q. A block of mass $2 k g$ rests on a horizontal surface. If a horizontal force of $5 N$ is applied on the block the frictional force on it is $(μ_{k} = 0.4, μ_{s} = 0.5)$

5 N

10 N

8 N

zero

Minimum force required to move the block
$= μ_{s} m g = 0.5 \times 2 \times 9.8 = 9.8 N$
Since the force applied is only $5 N$ , the block fails to move.
$∴$ Frictional force $=$ applied force $= 5 N$

Q. A block of mass 2kg rests on a horizontal surface. If a horizontal force of 5N is applied on the block the frictional force on it is (μk​=0.4,μs​=0.5)

5 N

10 N

8 N

zero

Minimum force required to move the block =μs​mg=0.5×2×9.8=9.8N Since the force applied is only 5N, the block fails to move. ∴ Frictional force = applied force =5N

Q. A block of mass $2 k g$ rests on a horizontal surface. If a horizontal force of $5 N$ is applied on the block the frictional force on it is $(μ_{k} = 0.4, μ_{s} = 0.5)$

Minimum force required to move the block
$= μ_{s} m g = 0.5 \times 2 \times 9.8 = 9.8 N$
Since the force applied is only $5 N$ , the block fails to move.
$∴$ Frictional force $=$ applied force $= 5 N$