Asked by Camille
A solid sphere of mass 0.604 kg rolls without slipping along a horizontal surface with a translational speed of 5.25 m/s. It comes to an incline that makes an angle of 38 with the horizontal surface. Neglecting energy losses due to friction,
(a) what is the total energy of the rolling sphere?
(b) to what vertical height above the horizontal surface does the sphere rise on the incline?
(a) what is the total energy of the rolling sphere?
(b) to what vertical height above the horizontal surface does the sphere rise on the incline?
Answers
Answered by
Elena
(a) KE= KE(transl) +KE(rot) = m•v²/2+Iω²/2=
= m•v²/2+ (2mR²)• v²/5•2•R²=
= m•v²/2+ mv²/5 =0.7 mv²
(b)
KE=PE
0.7•m•v²=mgh
h= 0.7•v²/g =…
= m•v²/2+ (2mR²)• v²/5•2•R²=
= m•v²/2+ mv²/5 =0.7 mv²
(b)
KE=PE
0.7•m•v²=mgh
h= 0.7•v²/g =…
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