Question

a block of mass 37kg restson a rough horizontal plane having coefficient of static friction 0.3. find out the least force required to just move the block horizontally​

Answer 1

Answer:force must be greater than 101 N bcz.

F> 0.3mg. To move it

F = 0.3×37×10

Explanation:

Answer 2

Answer:

Answer should be 111 N or 108.78 N

Explanation:

Coefficient of Friction (u)= 0.3

Mass of the body = 37 kg.

g = 10 m/s^2 or 9.8 m/s^2.

Force in the direction of y = 0

Therefore, N = mg

Here if we consider g as 10 then we get N = 370 N.

And if you consider g as 9.8 then we get N = 362.6 N.

Now as we know the formula of Friction,

Fr = uN

= 370 X 0.3 / 362.6 X 0.3

= 111 N / 108.78 N.

Therefore least value of Force should be 111 N or 108.78 N.