A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 4.01 m/s at the bottom of the rise. Find the translational speed at the top. The diagram illustrates a rolling motion of a sphere on an inclined surface. The sphere is depicted at two positions: 1. On the left, the sphere is shown moving towards an incline, with red arrows indicating both its linear motion to the right and its rotational motion in a clockwise direction.2. On the right, the sphere has climbed the incline and is at a higher height. The linear and rotational movements are similarly depicted with red arrows.The height of the incline is labeled as 0.760 meters. This measurement is marked vertically from the base level to the top of the incline where the sphere reaches its highest point. This diagram helps visualize concepts of energy conservation, motion, and dynamics in physics, specifically focusing on how both translational and rotational energies contribute to motion on an inclined surface.

using energy conservation energy at bottom=energy at top energy at bottom=potential energy+ kinetic energy=0+(translational kinetic energy+ rotational kinetic energy) potential energy at bottom is zero kinetic energy at bottom=12mv2+12IωB2 (1) m is mass, vB is translational speed at bottom, ωB is angular speed at bottom I is moment of inertia ball is solid sphere so I about center of mass is 25mr2 (r is radius of ball) ωb=vbrput these in eq(1) kinetic energy at bottom=12mvb2+12(25mr2)(vbr)2=12mvb2+15mvb2kinetic energy at bottom=710mvb2 energy at top =potential energy +kinetic energy energy at top=mgh+710mvt2 vt is translational speed at top, h is heightenergy at bottom=energy at top 0+710mvb2=mgh+710mvt2rearranging for vtvt=vb2-107gh put vb=4.01 h=0.760m g=9.8 m/s2 vt=4.012-(107×9.8×0.760)vt=2.332402 m/s translational speed at top is 2.332402 m/s (answer)