Question

A 12 kg rests on a horizontal surface and the coefficient of friction is 0.15. What is the smallest force that can give the mass an acceleration of 3 m/sec^2.

Answer 1

if we use g 10m/sec^2 then F=18Newton

Answer 2

Given:

A 12 kg rests on a horizontal surface and the coefficient of friction is 0.15.

To Find:

The minimum force, $F$ that can give the mass an acceleration of $3m/s^2$.

Solution:

Here, the coefficient of friction, μ$_s=0.15$

Mass of the body, $m=12kg$

Weight,

$W=mg=12*10=120N$     (Taking acceleration due to gravity, $g=10m/s^2$)

Since there is no vertical motion in the mass, this means that the weight is balanced by the normal reaction of the surface

∴ Normal reaction, $N=W=120N$

Limiting friction, $f_s=$ μ$_sN$

                      ⇒ $f_s=18N$

For producing an acceleration $3m/s^2$ , the net force required $=ma$

                                                                                                  $=12*3=36N$

    $F-f_s=36N$

⇒ $F-18=36N$

⇒ $F=54N$

Hence, the smallest force that can give the mass acceleration   $3m/s^2$ is 54N.