What moon phase occurs 3-4 days after a waning gibbous?
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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.
- The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
- The height at which the block stops rising is approximately 1.1415 m
- The length of the incline is approximately 1.536 m
Mass of the block, m = 3 kg
Coefficient of kinetic friction, = 0.39
Therefore, we have;
Friction force = ·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, <em>W</em> ≈ 7.68 × d
The work done = The kinetic energy of the block, which gives;
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, <em>W</em> ≈ 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 velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>
The height at which the block will stop moving, <em>h</em>, is given as follows;
Which gives;
The distance up the inclined, the block rises, at maximum height is therefore;
≈ 1.536 m
Therefore;
h = 1.536 × sin(48°) ≈ 1.1415
- The height at which the block stops rising, h ≈ <u>1.1415 m</u>
From the above solution for the height, the length of the incline is he
distance along the incline at maximum height which is therefore;
- Length of the incline,
= 1.536 m
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