A solid sphere of mass 2.5 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15° and rolls to the bottom.The upper end of the ramp is 1.6 m higher than the lower end.What is the linear speed of the sphere when it reaches the bottom of the ramp? (Note: I = 0.4MR² for a solid sphere,and g = 9.8 m/s². ) A) 4.7 m/s B) 22.4 m/s C) 5.6 m/s D) 3.3 m/s

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Explanation:The potential energy at the top of the ramp is:PE = mgh = (2.5 kg)(9.8 m/s^2)(1.6 m) = 39.2 JThe kinetic energy at the bottom of the ramp is:KE = 1/2mv^2The work done by gravity is equal to the change in kinetic energy:W = KE - PE = 1/2mv^2 - 39.2 JThe work done by gravity is also equal to:W = mgh = (2.5 kg)(9.8 m/s^2)(1.6 m) = 39.2 JTherefore,1/2mv^2 - 39.2 J = 39.2 JSolving for v, we get:v = sqrt(2(39.2 J) / 2.5 kg) = 4.7 m/s

A Solid Sphere of Mass 2