Physics, 06.07.2019 05:30 krizzle168

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?

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Answer from: misslux

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;

F = mg sin θ

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

The frictional force F = μN

where;

N = mg cos θ

∴

The frictional force F = μ × mg cos θ

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

F = mg sinθ - μ mg cos θF = sin θ - μ cos θ

Similarly, using the second equation of motion

Force of an object = mass × acceleration

∴

Total force = mg cos θ - μmg cos θma = mg cos θ - μmg cos θ

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

acceleration (a) = g(sin θ - μ cos θ)

a = 9.8 × (sin 14 - 0.380 ×cos 14)a = 9.8 × (0.2419 - 0.380 × 0.9703)a = 9.8 × (0.2419 - 0.368714)a = - 1.2427 m/s²

Finally, using the kinematics equation;

$\mathbf{v^2 - u^2 =2as}$

where;

v = final velocityu = initial velocitys = distance of the boxa = acceleration

$\mathbf{v^2=u^2+2as }$

$\mathbf{(0)^2=(1.70)^2+2\times (-1.2427)s }$

$\mathbf{-(1.70)^2=+2\times (-1.2427)s }$

Making (s) the subject of the formula:

$\mathbf{s = \dfrac{-(1.70)^2}{2 \times -1.2427}}$

$\mathbf{s = \dfrac{(1.70)^2}{2 \times 1.2427}}$

$\mathbf{s = \dfrac{2.89}{2.4854}}$

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:

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Answer from: graciewyatt471

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

F_x = mg sin14

F_y = mg cos14

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

F_n = mg cos14

now to find the friction force we will use

$F_f = \mu * F_n$

F_f = 0.380 * mg cos14

Now the net force along the inclined plane is given as

$F_{net} = mgsin14 - F_f$

$F_{net} = mgsin14 - 0.380 mgcos14$

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

$a = \frac{F_{net}}{m}$

a = gsin14 - 0.380gcos14

a = 9.8(sin14 - 0.380*cos14)

a = -1.24 m/s^2

now by using kinematics we will write

v_f^2 - v_i^2 = 2 a d

0 - 1.70^2 = 2*(-1.24)*d

d = 1.16 m

so it will cover the distance of 1.16 m

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