The ball's horizontal and vertical velocities at time are
but the ball is thrown horizontally, so . Its horizontal and vertical positions at time
are
The ball travels 22 m horizontally from where it was thrown, so
from which we find the time it takes for the ball to land on the ground is
When it lands, and
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
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 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
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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