A block of mass 1kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5m/s2, the frictional force acting on the block is,
[Take g=10m/s2]
A
6N
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B
10N
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C
15N
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D
5N
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Solution
The correct option is D5N Pseudo force method:
Here, N=mg=1×10=10N ⇒fsmax=μ×N=0.6×10=6N
Now, Fpseudo=ma=1×5=5N
As, Fpseudo<fsmax
∴The required frictional force is of 5N magnitude.