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A 1.00 kg sphere M, suspended by a string from point P, is lifted to a height h. The sphere is released and passes through the lowest point in its swing at a speed of 10.0 meters per second. [Neglect friction.]
a. Calculate the height from which the sphere was released. [Show all work including equation and substitution with units]

b. Compared to the sphere’s speed through the lowest point of its swing when released from h, the sphere’s speed through the lowest point when released from 2h would be

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hamzaahmeds

Answer:

(a) h = 5.1 m

(b) v = 14.13 m/s

Explanation:

(a)

We will use the law of conservation of energy. For this situation it states that:

Loss in Potential Energy of Sphere = Gain in Kinetic Energy of the Sphere

mgh = (1/2)mv²

h = v²/2g   --------------- equation (1)

where,

h = height = ?

v = speed at lowest point = 10 m/s

g = 9.8 m/s²

Therefore,

h = (10 m/s)²/(2)(9.8 m/s²)

h = 5.1 m

(b)

using the equation (1)

h = v²/2g

v = √2gh

where,

v = velocity = ?

g = 9.8 m/s²

h = height = 2(5.1 m) = 10.2 m

Therefore,

v = √[2(9.8 m/s²)(10.2 m)]

v = 14.13 m/s

onyebuchinnaji

The height from which the sphere was released is 5.1 m

When the height is doubled, the speed of the sphere becomes 14.14 m/s.

The given parameters:

  • Mass of the sphere, m = 1.0 kg
  • Speed of the sphere, v = 10 m/s

The height from which the sphere was released is calculated as follows;

[tex]P.E = K.E\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2}v^2\\\\h = \frac{v^2}{2g}\\\\h = \frac{(10)^2}{2(9.8)} \\\\h = 5.1 \ m[/tex]

When the height is doubled, the speed of the sphere becomes;

[tex]h = \frac{v^2}{2g} \\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 9.8 \times (2 \times 5.1)} \\\\v = 14.14 \ m/s[/tex]

Learn more about  conservation of mechanical energy here: https://brainly.com/question/6852965

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