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A uniform solid sphere with a mass of M = 390 grams and a radius R = 19.0 cm is rolling without slipping on a horizontal surface at a constant speed of 4.00 m/s. It then encounters a ramp inclined at an angle of 15.0 degrees with the horizontal, and proceeds to roll without slipping up the ramp. Use g = 10.0 m/s2. How far does the sphere travel up the ramp (measure the distance traveled along the incline) before it stops for an instant?

Respuesta :

Answer:

l = 4.33 m

Explanation:

given,

mass of solid sphere = 390 gram = 0.39 kg

radius = R = 19 cm = 0.19 m

rolling with constant speed = 4 m/s

angle with horizontal = 15°

acceleration due to gravity = 10 m/s²

using energy conservation

[tex]\dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2 = mgh[/tex]

I for sphere

[tex]I = \dfrac{2}{5}mr^2[/tex]         v = r ω

[tex]\dfrac{1}{2}\ \dfrac{2}{5}mr^2\times \dfrac{v^2}{r^2} + \dfrac{1}{2}mv^2 = mgh[/tex]

[tex]\dfrac{7}{10}mv^2 = mgh[/tex]

[tex]h = \dfrac{0.7 v^2}{g}[/tex]

[tex]h = \dfrac{0.7\times 4^2}{10}[/tex]

h = 1.12 m

[tex]l=\dfrac{h}{sin 15^0}[/tex]

[tex]l=\dfrac{1.12}{sin 15^0}[/tex]

l = 4.33 m

the sphere will travel 4.33 m on the ramp.

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