A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
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For rolling without slipping, use both translational and rotational equations to solve for acceleration.
Given values:- Applied Force: \( F = 49 \, \text{N} \)- Sphere Mass: \( m = 20 \, \text{kg} \)- Rolling without slipping condition is met.The condition for rolling without slipping is \( a = \alpha r \), where \( a \) is linear acceleration, \( \alpha \) is angular acceleration, and \( r \) is the sphere's radius.The applied force \( F \) drives both translation and rotation. The translational equation of motion is \( F = ma \).For rotational motion, torque \( \tau = I \alpha \). The moment of inertia for a solid sphere is \( I = \frac{2}{5}mr^2 \).Substituting these into the torque equation and using the no-slip condition: \( F \cdot r = \frac{2}{5}mr^2 \cdot \alpha \), which simplifies to \( F = \frac{2}{5}m \cdot a \).Solving for acceleration: \( a = \frac{5F}{2m} = \frac{5 \times 49}{2 \times 20} = 2.5 \, \text{m/s}^2 \).Thus, the sphere's center accelerates at \( 2.5 \, \text{m/s}^2 \).
