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/s^2) F = 28 N
a) 50
b) 30
c) 22
d) Zero​

Answer 1

Answer:

Here , maximum value of static friction that can act on the body = μ

s

mg

= 25 N

(i) Since the applied force is less than the maximum possible value of static friction

therefore , the force of friction will be just enough to stop the motion of the body.

Here , friction force = 15 N

(ii) Here, applied force = maximum possible value of static friction.

Friction force = 25 N

(iii) Here , applied force > maximum value of static friction.

Therefore , the body moves and kinetic friction acts on the body.

Friction force = μ

k

mg = 20 N.

Answer 2

(b)30

Explanation:

Formula :

$f = umg$

Given:

$m = 5kg$

$frictional force = 0.6$

The Frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. The main purpose of a frictional force is to create resistance to the motion of one surface over the other surface.

$g = 10 m/s^2$

Force of friction, $f = umg$

$m = mass\\u = frictional force \\g = gravitational force$

$f =0.6* 5*10$

$f=30N$