Answer :
The total kinetic energy of a 5 kg solid sphere with a diameter of 0.1 m rolling down an inclined plane at 4 m/s is 56 J.
The student asked about the total kinetic energy of a solid sphere rolling down an inclined plane. The sphere has a diameter of 0.1 m, mass of 5 kg, and is moving with a speed of 4 m/s. The total kinetic energy of an object is the sum of its translational kinetic energy and its rotational kinetic energy. For a solid sphere, the moment of inertia (I) is (2/5)mr2. The translational kinetic energy (Kt) can be calculated as (1/2)mv2, and the rotational kinetic energy (Kr) as (1/2)Iω2. Since ω can be expressed as v/r for a rolling object without slipping, we can substitute ω with v/r in the formula for Kr.
Calculating the translational kinetic energy:
Kt = (1/2) * m * v2 = (1/2) * 5 kg * (4 m/s)2 = 40 J
Calculating the rotational kinetic energy:
Kr = (1/2) * I * ω2 = (1/2) * [(2/5) * 5 kg * (0.05 m)2] * (4 m/s / 0.05 m)2 = 16 J
The total kinetic energy is the sum of translational and rotational kinetic energy:
KEtotal = Kt + Kr = 40 J + 16 J = 56 J
