A 10 kg box slides down a lane inclined at an angle θ = 30 . The plane has a friction o
coefficient 0.1. The box starts from the rest and slides down the plane for 2.0 s. What is
the distance that the box travels down the plane?

The distance that the box travels down the plane is 8.1 m.

The given parameters;

The acceleration of the box is calculated as follows;

The distance traveled by the box down the plane is calculated as follows;

Thus, the distance that the box travels down the plane is 8.1 m.

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The distance the box traveled down the plane is 19.28 m

Explanation:

The angle of repose, α, is given by the relation;

tan⁻¹(μ) = α

tan⁻¹(0.1) = 5.7°

Therefore, we have;

M·g·sin(θ) - μ·N = M·a

Where:

M = Mass of the box = 10 kg

g = Acceleration due to gravity = 9.81 m/s²

θ = Angle of inclination of the plane = 30°°

μ = Coefficient of friction = 0.1

a = Acceleration of the box along the incline plane

N = Normal force due to the weight of the box = M·g·cos(θ)

10 × 9.81 × sin30 - 0.1 × 9.81 × cos(30)  = 10 × a

48.2 = 10 ×a

a = 48.2/10 = 4.82 m/s²

The distance, s, traveled by the box is given by the relation;

s = u·t + 1/2×a·t²

Where:

u = Initial velocity = 0 m/s

t = Time of motion = 2.0 s

∴ s = 0×2 + 1/2 × 4.8 × 2² = 19.28 m

The box traveled 19.28 m down the plane.

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west