To solve the problem of finding the height attained by a solid sphere rolling up an inclined plane, we can use the work-energy theorem. Here’s a step-by-step solution:
### Step 1: Understand the Kinetic Energy of the Sphere
The total kinetic energy (KE) of the sphere consists of two parts: translational kinetic energy and rotational kinetic energy.
- **Translational Kinetic Energy (TKE)**:
\[
\text{TKE} = \frac{1}{2} mv^2
\]
- **Rotational Kinetic Energy (RKE)**:
\[
\text{RKE} = \frac{1}{2} I \omega^2
\]
Where:
- \( m \) is the mass of the sphere (2 kg),
- \( v \) is the linear velocity (10 m/s),
- \( I \) is the moment of inertia of the sphere, and
- \( \omega \) is the angular velocity.
### Step 2: Calculate the Moment of Inertia
For a solid sphere, the moment of inertia \( I \) is given by:
\[
I = \frac{2}{5} m r^2
\]
However, we will use the relationship between linear velocity and angular velocity:
\[
\omega = \frac{v}{r}
\]
### Step 3: Substitute \( \omega \) into the RKE Formula
Substituting \( \omega \) into the RKE formula gives:
\[
\text{RKE} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{5} mv^2
\]
### Step 4: Total Kinetic Energy
Now, we can write the total kinetic energy:
\[
\text{Total KE} = \text{TKE} + \text{RKE} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2
\]
To combine these, we need a common denominator:
\[
\text{Total KE} = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2
\]
### Step 5: Apply the Work-Energy Theorem
According to the work-energy theorem, the total kinetic energy will be converted into potential energy (PE) when the sphere reaches its maximum height:
\[
\text{Total KE} = mgh
\]
Where \( h \) is the height attained, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
### Step 6: Set Up the Equation
Now, we can set up the equation:
\[
\frac{7}{10} mv^2 = mgh
\]
### Step 7: Cancel Mass and Solve for Height
Since the mass \( m \) appears on both sides, we can cancel it out:
\[
\frac{7}{10} v^2 = gh
\]
Now, substituting \( v = 10 \, \text{m/s} \) and \( g = 9.81 \, \text{m/s}^2 \):
\[
\frac{7}{10} (10)^2 = 9.81h
\]
\[
\frac{7}{10} \times 100 = 9.81h
\]
\[
70 = 9.81h
\]
Now, solve for \( h \):
\[
h = \frac{70}{9.81} \approx 7.13 \, \text{m}
\]
### Final Answer
The height attained by the sphere before it stops is approximately \( 7.13 \, \text{m} \).
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