a person throws a ball in such a way that its speed is zero at one particular point in its path. How did the person throw the ba
ll?
1 answer:
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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
Learn more about conservation of energy here:
Answer:
F = 47.6 N
Explanation:
- Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:
- So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:
⇒ Fsk = 47.6 N (normal to the wall)