Question

A block with a mass of 5.0 kg slides down a 370 incline as 1 point

shown below with an acceleration of 5.6 m/s2 (sin 370 =
0.6, cos 370 = 0.8). The coefficient of kinetic friction
between the block and the inclined surface is 0.050. The
magnitude of the friction force along the plane is nearly: *
2N
5N
6N
30N

Answer 1

2 N

Step-by-step explanation:

The equation of the forces along the direction perpendicular to the plane is

`R-mg cos \theta = 0`

where

R is the reaction force

`mg cos \theta` is the component of the weight perpendicular to the plane, with

m = 5.0 kg being the mass of the plane

`g=9.8 m/s^2` acceleration of gravity

`\theta=37^(\circ)`

Solving for R,

`R=mgcos \theta = (5.0)(9.8)(0.8)=39.2 N` (1)

The frictional force is given by

`F=\mu R`

where

`\mu=0.050` being the coefficient of friction

Using (1), we find

`F=(0.050)(39.2)=2 N`