A block of mass m=1.8 kg slides along the inside of a hemispherical bowl of radi

Question

A block of mass m=1.8 kg slides along the inside of a hemispherical bowl of radius r=0.8 m. When the block is at an angle of 65 degrees above the lowest point it is sliding down with a speed of 3 m/s and is speeding up at a rate of 5 m/s^2. At this point find a.) the normal force on the block by the bowl, b.) the kinetic friction force on the block by the bowl and c.) the coefficient of friction between the block and the bowl.

Explanation / Answer

(a) the normal force on the block by the bowl = Mgsin? + Mv^2/r = 1.8 * 9.8 * sin 25 + 1.8 * (3)^2 / 0.8 = 27.7 N (b) Mgcos? - ?(Mgsin? + Mv^2/r) = Ma ===> the kinetic friction force on the block by the bowl = Mgcos? - Ma = 1.8 * 9.8 * cos 25 - 1.8 * 5 = 6.987 N (c) ?(Mgsin? + Mv^2/r) = 6.987 N ===> ? = 6.987 / 27.7 = 0.25

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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