Question

A 10 kg box slides down a lane inclined at an angle θ = 30 . The plane has a friction of 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?

Answer 1

Answer:

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.

Answer 2

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

The given parameters;

• mass, m = 10 kg
• angle, θ = 30⁰
• coefficient of friction, μ = 0.1
• time of motion of the box, t = 2 s

The acceleration of the box is calculated as follows;

$\Sigma F_{net} = ma\\\\mgsin\theta - \mu mg cos\theta = ma\\\\gsin\theta - \mu g cos\theta = a\\\\g(sin\theta - \mu cos \theta ) = a\\\\9.8(sin30 \ - \ 0.1\times cos 30) = a\\\\9.8(0.413) = a\\\\4.05 \ m/s^2 = a$

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

$s= ut + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} \times 4.05 \times 2^2\\\\s = 8.1 \ m$

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

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