Question

L../IN10. An object of mass 60.986 kg is sliding on a horizontalsurface with a uniform speed. The coefficientof kinetic fiction of the surfaces is 0.15. Calculatethe force of friction exerted by the surface on the object. (1 point)A. O 101.798 NB. O114.077 NC. 050.2 ND. 145.451 NE. O 89.649 N

Answer 1

Given:

The mass of the object is,

`m=60.986\text{ kg}`

The coefficient of friction is,

`\mu=0.15`

The object is sliding on the horizontal surface at a uniform speed.

To find:

the force of friction exerted by the surface on the object

Step-by-step explanation:

If we draw the free body diagram of the object we see,

The normal reaction is R, the applied force is F, and the frictional force isf.

As the object is continuing its constant speed, we can say the object is in equilibrium.

So,

`R=mg`

and

`\begin{gathered} f=F=\mu R \\ f=\mu mg \\ f=0.15*60.986*9.8 \\ f=89.649\text{ N} \end{gathered}`

Hence, the friction force exerted by the surface is 89.649 N.