Abox is sliding down an incline tilted at an angle 14.0° above horizontal. the box is sliding down the incline at a speed of 1.70 m/s. the coefficient of kinetic friction between the box and the incline is 0.380. how far does the box slide down the incline before coming to rest?

A box is sliding down an incline tilted at an angle 14.0° above horizontal. The box is sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.380. The box slide at 1.16 m down the incline before coming to rest.

The diagrammatic expression of the question given can be seen in the image below.

We need to resolve the component of the plane into parallel and perpendicular forces.

In the parallel direction;

In the perpendicular direction, the normal force N = mg cos θ

The frictional force F = μN

where;

∴

The frictional force F = μ × mg cos θ

Thus, the resultant force in the parallel direction of the plane is:

Similarly, using the second equation of motion

Force of an object = mass × acceleration

∴

Making acceleration (a) the subject of the formula, we have:

acceleration (a) = g(sin θ - μ cos θ)

Finally, using the kinematics equation;

where;

Making (s) the subject of the formula:

s = 1.16 m

Therefore, we can conclude that the distance of how far the box slides down the incline before coming to rest is 1.16 m

Learn more about kinetic friction here:

link

The two components of the weight of the block is given as

now the normal force on the block will be counter balanced by Y component of weight of block

now to find the friction force we will use

Now the net force along the inclined plane is given as

now the acceleration of block is given by Newton's II law

now by using kinematics we will write

so it will cover the distance of 1.16 m

6/12=1/2

step-by-step explanation:

this could be newton#'s law of cooling. the cooling curve is exponential