Question

A 177-Newton horizontal force is exerted upon a 6.4-kg box to move it across a level surface at a constant velocity of 2.1 m/s.The force of friction encountered by the box is

Answer 1

62.72 N

Step-by-step explanation

Step 1

Free body diagram

so.F.b.d

the FBD shows us that the block is in equilibrium in the vertical direction, so

`\begin{gathered} N-mg=o \\ N=mg \\ N=6.4\text{ kg*9.8}(m)/(s^2) \\ N=62.72\text{ N} \end{gathered}`

The normal contact force applied by the surface is 62.72 N,thus by Newton's third law of motion, 62.72 N force will be exerted by the block on the surface.

therefore, we can conclude

the exerted force is 62.72 N

I hope this helps you