Answer:
0.231 m
Explanation:
Given that
Initial velocity, u = 1.4 m/s
Coefficient if friction, μ = 0.180
Final velocity, v = 0 m/s
Since the angle made with the inclined isn't stated. I'd be working with an angle of 15°.
The formula for the acceleration is given as
Acceleration = f.r.m.g.sinθ/m
Where f = μN = μ.m.g.cosθ
On substituting, acceleration a then is equal to
a = -(u.g.cosθ + g.sinθ)
a = -(0.18 * 9.8 * cos15 + 9.8 sin15)
a = -(1.764 * 0.9659 + 9.8 * 0.2588)
a = -(1.7038 + 2.5362)
a = -4.24 m/s²
To find the distance traveled, we use one of the equations of motions
v² = u² + 2as, with v = 0
u² = -2as
s = -u²/2a, on substituting
s = -1.4² / -2 * 4.24
s = 1.96 / 8.48
s = 0.231 m
The distance traveled by the box before coming to rest is 0.23 m.
The given parameters;
The normal force on the box is calculated as follows;
[tex]F_n = mgsin\ \theta[/tex]
The frictional force on the box, is calculated as follows;
[tex]F_f = \mu_k F_n\\\\[/tex]
The net force on the box is calculated as follows;
[tex]-mgsin \ \theta - \mu F_n = ma\\\\-m(gsin\theta + \mu gcos\theta) = ma\\\\- g(sin\theta + \mu cos\theta) = a\\\\-9.8(sin \ 15 \ + \ 0.18\times cos\ 15) = a\\\\-4.24 \ m/s^2 = a[/tex]
The distance traveled by the box before coming to rest is calculated as;
[tex]v^2 = u^2 + 2ad\\\\0 = u^2 + 2ad\\\\0 = (1.4)^2 + 2(-4.24)d\\\\0 = 1.96 - 8.48d\\\\8.48d = 1.96\\\\d = \frac{1.96}{8.48} \\\\d = 0.23\ m[/tex]
Thus, the distance traveled by the box before coming to rest is 0.23 m.
"Your question is not complete, it seems to be missing the following information";
A box is sliding up an incline that makes an angle of 15 degrees with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.40 m/s. How far does the box travel along the incline before coming to rest.
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