Question

the force of friction on the block at t = 12 second
is
Weight of block: 5kg
Value of mhu: 0.1
Applied force: 0.5t ​

Answer 1

Answer:

friction force = u × N

mass of block =5kg

so N = 50N

so , Fr ( fictional force ) = 0.1 × 50

= 5N

Answer 2

Answer:

The value of frictional force is 4.9 N.

Explanation:

• Applied force: The applied force is, F = 0.5t

           At t = 12 second, the force = F = 0.5 × 12 = 6 N.

• Force of Friction: The force of friction = $F_friction$ = μN

        μ = coefficient of friction = 0.1

        N = Normal force = mg = 5 × 9.8 = 49 N

       Force of friction =  $F_friction$ = μN = 0.1 × 49 N = 4.9 N

      The frictional force is less than the applied force, so the block will move.

     The value of frictional force is 4.9 N.

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