Respuesta :
Answer:
1/2
Explanation:
We need to make a couple of considerations but basically the problem is solved through the conservation of energy.
I attached a diagram for the two surfaces and begin to make the necessary considerations.
Rough Surface,
We know that force is equal to,
[tex]F_r = mgsin\theta[/tex]
[tex]F_r = \mu N[/tex]
[tex]F_r = \mu mg cos\theta[/tex]
Matching the two equation we have,
[tex]\mu N = \mu mg cos\theta[/tex]
[tex]\mu = tan\theta[/tex]
Applying energy conservation,
[tex]\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1[/tex]
[tex]\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1[/tex]
[tex]\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1[/tex]
[tex]\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1[/tex]
[tex]h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})[/tex]
Frictionless surface
[tex]\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2[/tex]
[tex]\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2[/tex]
[tex]\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2[/tex]
[tex]h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})[/tex]
Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between [tex]h_1[/tex] and [tex]h_2[/tex] is given by
[tex]\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}[/tex]
[tex]\frac{h_1}{h_2} = \frac{1}{2}[/tex]
The presence of friction enables the the ball to roll higher up the hill by
converting its rotational kinetic energy into potential energy.
[tex]\underline{\mathrm{The \ ratio \ of \ \dfrac{h_1}{h_2} \ is \ \dfrac{7}{5}}}[/tex]
Reasons:
The initial kinetic energy of each ball are given as follows;
Work done by ball rolling up a hill without slipping.
For the bowling ball that rolls up the hill without slipping, we have;
[tex]\dfrac{1}{2} \cdot (c + 1) \cdot M \cdot v_0^2 = \mathbf{M \cdot g \cdot h}[/tex]
Where, for a sphere, [tex]c = \dfrac{2}{5}[/tex]
Therefore;
[tex]\dfrac{1}{2} \cdot \left(\dfrac{2}{5} + 1 \right) \cdot M \cdot v_0^2 = M \cdot g \cdot h_1[/tex]
[tex]\mathbf{\dfrac{7}{10} \cdot v_0^2} = g \cdot h_1[/tex]
[tex]h_1 = \mathbf{\dfrac{7}{10} \cdot\dfrac{ v_0^2 }{g}}[/tex]
For the bowling hall that slides up the hill, we have;
[tex]\dfrac{1}{2} \cdot M \cdot v_0^2 = \mathbf{M \cdot g \cdot h_2}[/tex]
[tex]\dfrac{1}{2} \cdot v_0^2 = g \cdot h_2[/tex]
[tex]h_2 = \mathbf{\dfrac{v_0^2 }{2 \cdot g} \cdot}[/tex]
[tex]\mathrm{The \ ratio \ of \ \dfrac{h_1}{h_2} = \mathbf{\frac{ \dfrac{7}{10} \cdot\dfrac{ v_0^2 }{g}}{\dfrac{v_0^2 }{2 \cdot g} \cdot} }} = \dfrac{7}{10} \times 2 = \dfrac{7}{5}[/tex]
[tex]\underline{\mathrm{The \ ratio \ of \ \dfrac{h_1}{h_2} = \dfrac{7}{5}}}[/tex]
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https://brainly.com/question/22210059
