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Question

Physics Question on laws of motion

A block of mass 2kg2\,kg rests on a horizontal surface. If a horizontal force of 5N5\,N is applied on the block the frictional force on, it is (μk=0.4,μs=0.5)(\mu _{k}=0.4,\,\mu _{s}=0.5)

A

5N5\,\,N

B

10N10\,\,N

C

8N8\,\,N

D

zero

Answer

8N8\,\,N

Explanation

Solution

The frictional force FF is given by
F=μkmgF=\mu _{k}mg
Given, μk=0.4, \mu_{k}=0.4,
m=2kgm=2\,kg,
g=10m/s2g=10 \, m/s^{2}
\therefore F=0.4×2×10F=0.4\times 2\times 10
F=8NF=8\,N