Question

A block of mass 5 kg is placed on a horizontal
surface with coefficient of friction u = 0.2, then the
maximum and minimum value of force F for which
the block remains at rest are (g = 10 m/s2)
15 N
5 kg
> F
(1) 15 N, 10 N
(3) 10 N, 25 N
(2) 25 N, 5 N
(4) 5 N, 25 N
If moss​

Answer 1

Answer:

25 N,5 N

Explanation:

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Answer 2

Answer:

Option (2)25 N, 5 N

Explanation:

We are given that  mass of  a block=5 kg

Coefficient of friction=0.2

We have to find  the maximum and minimum value of force F for which the block remains at rest.

We are given that $g=10m/s^2$

Friction force=$\mu mg=0.2\times 5\times 10=10 N$

If friction force applied in the direction of force 15 N

Then , Net force=f+F=10+15=25 N

If friction force applied in opposite direction to 15 N then

Net force=F-f=15-10=5 N

Hence, maximum force =25 N

Minimum force=5 N

Option (2)25 N, 5 N