Question

A body of mass 5 kg is kept on a rough horizontal floor of coefficient of static friction us = 0.6. A force F is applied on the body as shown in figure. Value of friction acting on the body is (g = 10 m/s2) → F = 28 N 7 7
a) 50 N
b) 30 N
c) 28 N
d) Zero​

Answer 1

Answer:

20 is the answer of the question

Answer 2

Answer: The value of the friction acting on the body is 30N.

Explanation: Given that,

$m = 5kg\\\mu_s = 0.6\\g = 10m/s^2\\F = 28N$

We know that

$f = \mu_sN$

where N is the normal reaction.

$f = \mu_smg\\f =0.6\times10\times5\\f = 30N$