A block of mass 1 kg is placed on a truck which accelerates with acceleration 5 m s- 2 . The coefficient of static friction between the block and truck is text0.6 . The frictional force acting on the block is

Question

A block of mass 1 kg is placed on a truck which accelerates with acceleration 5 m s- 2 . The coefficient of static friction between the block and truck is text0.6 . The frictional force acting on the block is (A) 5N (B) 6N (C) 5.82N (D) 4.6N

Tardigrade Admin · Accepted Answer

Given, a=5ms- 2,m=1 kg,μ = text0.6 . ∴ Maximum static frictional force fs=μ smg=0.6× 1× 10=6N Force on a block due to acceleration of truck Fext=ma=1× 5=5N Hence, Fext < fs Therefore, force of friction = external force = 5N

Q. A block of mass $1 k g$ is placed on a truck which accelerates with acceleration $5 m s^{- 2}$ . The coefficient of static friction between the block and truck is $0.6$ . The frictional force acting on the block is

$5 N$

$6 N$

$5.82 N$

$4.6 N$

Given, $a = 5 m s^{- 2}, m = 1 k g, μ = 0.6$ .
$∴$ Maximum static frictional force $f_{s} = μ_{s} m g = 0.6 \times 1 \times 10 = 6 N$
Force on a block due to acceleration of truck $F_{e x t} = ma = 1 \times 5 = 5 N$
Hence, $F_{e x t} < f_{s}$
Therefore, force of friction = external force = $5 N$

Q. A block of mass 1kg is placed on a truck which accelerates with acceleration 5ms−2 . The coefficient of static friction between the block and truck is 0.6 . The frictional force acting on the block is

5N

6N

5.82N

4.6N

Given, a=5ms−2,m=1kg,μ=0.6 . ∴ Maximum static frictional force fs​=μs​mg=0.6×1×10=6N Force on a block due to acceleration of truck Fext​=ma=1×5=5N Hence, Fext​<fs​ Therefore, force of friction = external force = 5N

Q. A block of mass $1 k g$ is placed on a truck which accelerates with acceleration $5 m s^{- 2}$ . The coefficient of static friction between the block and truck is $0.6$ . The frictional force acting on the block is

$5 N$

$6 N$

$5.82 N$

$4.6 N$