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Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Determine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)

Respuesta :

Numenius

Answer:

The maximum height is 0.33 m.

Explanation:

initial velocity, u = 8 m/s

final velocity, v = 0 m/s

10% of  kinetic energy is lost in friction.

The kinetic energy used to move up the top,

KE = 10 % of 0.5 mv^2

KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m

Let the maximum height is h.

Use conservation of energy

KE at the bottom = PE at the top

3.2 m = m x 9.8 x h

h = 0.33 m  

Lanuel

The height traveled vertically up the hill by the ball when it stops is 0.327 meter.

Given the following data:

  • Velocity = 8.0 m/s
  • Kinetic energy = 10% lost to friction.

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

To determine how far (height) vertically up the hill the ball reaches when it stops:

By applying the law of conservation of energy, we have:

Kinetic energy lost at the bottom = Potential energy gained at the top.

Mathematically, the above expression is given by the formula:

[tex]0.1 \times \frac{1}{2} mv^2 = mgh\\\\0.1 \times \frac{1}{2} v^2 = gh\\\\h=\frac{0.1v^2}{2g}[/tex]

Substituting the given parameters into the formula, we have;

[tex]h=\frac{0.1 \times 8^2}{2\times 9.8} \\\\h=\frac{0.1 \times 64}{19.6} \\\\h=\frac{6.4}{19.6}[/tex]

Height, h = 0.327 meter.

Read more on kinetic energy here: https://brainly.com/question/17081653

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