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

Answer:

2 N

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

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