A 2.12-kg solid sphere (radius 0.126 m) is released from rest at the top of a ra

Question

A 2.12-kg solid sphere (radius 0.126 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.764 m high and 5.22 m long. When the sphere reaches the bottom of the ramp, what is its total kinetic energy? (in ) A: 5.21B: 7.56C:1.10x10D: 1.59x 10 E: 2.30x10F:3.34x101G: 4.84x101 H: 7.02x10 Submlt An Tries 0/3 Whet is its rotational kinetic energy? (in 1) A: 4.54|| B: 5.31 C: 6.21|| D: 727| E: 8.51 | F: 9.95| G: 1.16x10 H:1.36x101 Submit Answer Tries 0/3 What is its translational kinetic energy? (in ?) A: 8.42x10-11 B: 1.2? C: 1.77 D: 2.57|| E: 3.72] F: S.40 G: 7.831 H: 1.13×101 Submit Arswer Tries 0/3

Explanation / Answer

a) From law of conservation of energy,

Total KE at bottom = Total PE at top

PE at top = mgh = (2.12)(9.8)( 0.764) = 15.87 J

Total KE at bottom= 15.87 J option (d)

c) using the conservationm of energy

mgh = 0.5mv^ 2 + 0.5I?2

I = 2mr^2/5,

mgh = 0.5mv^2 + 0.5(2/5mr^2)( v^2/r^2)

gh = 0.5v^2 + 1/5v^2

gh= 7/10v^2

v = sqrt( 10gh/7)

v = sqrt( 10x 9.8 x 0.764/7)= 3.27 m/s approx.

Translational KE at bottom = 0.5mv^2 = 0.5(2.12)(10.70)^2 = 11.34 J

b) rotational KE at bottom= 15.87 - 11.34 = 4.53 J

Part a) Option (d)

Part b) Option (a)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.