A6.0-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

a.

Explanation:

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

- Along the parallel direction:

where :

m = 6.0 kg is the mass  of the box

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

 is the angle of the slope

is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

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

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

Solving for a,

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards