Question

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic friction is 0.40, at what rate does the box accelerate down the slope?
a. 3.1 m/s2
b. 3.4 m/s2
c. 3.7 m/s2
d. 4.1 m/s2

Answer 1

Answer:

a. $3.1 m/s^2$

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

$mg sin \theta - \mu_k R = ma$

where :

m = 6.0 kg is the mass  of the box

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

$\theta=39^{\circ}$  is the angle of the slope

$\mu_k = 0.40$ is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

$R-mg cos \theta =0$

From the 2nd equation, we get an expression for the reaction force:

$R=mg cos \theta$

And substituting into the 1st equation, we can find the acceleration:

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2$