The compression of 10 cm by a 100 N force on the plane that has a
coefficient of friction of 0.39 give the following values.
Mass of the block, m = 3 kg
[tex]Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}[/tex]
Coefficient of kinetic friction, [tex]\mu_k[/tex] = 0.39
Therefore, we have;
Friction force = [tex]\mathbf{\mu_k}[/tex]·m·g·cos(θ)
Which gives;
Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68
Work done by the motion of the block, W ≈ 7.68 × d
The work done = The kinetic energy of the block, which gives;
[tex]\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d[/tex]
The initial kinetic energy in the spring is found as follows;
K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J
The initial velocity of the block is therefore;
5 = 0.5·m·v²
v₁ = √(2 × 5 ÷ 3) ≈ 1.83
Work done by the motion of the block, W ≈ 7.68 N × 0.07 m ≈ 0.5376 J
Chane in kinetic energy, ΔK.E. = Work done
ΔK.E. = 0.5 × 3 × (v₁² - v₂²)
Which gives;
ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376
Which gives;
The height at which the block will stop moving, h, is given as follows;
[tex]At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x[/tex]
Which gives;
[tex]Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536[/tex]
The distance up the inclined, the block rises, at maximum height is therefore;
[tex]x_{max}[/tex] ≈ 1.536 m
Therefore;
h = 1.536 × sin(48°) ≈ 1.1415
From the above solution for the height, the length of the incline is he
distance along the incline at maximum height which is therefore;
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