A 40-kg girl swings from a rope, starling at rest at height h. When she reaches

Question

A 40-kg girl swings from a rope, starling at rest at height h. When she reaches the lowest point in her arc, she releases the rope and lands on top of a 10-kg sled sitting at rest along ice-covered ground. The girl and the sled slide together along the ground a distance 15 m before coming to a complete stop. The coefficient of kinetic friction between the sled and the ground is 0.12. To solve this problem, we need to work it backwards. Use energy to find the speed of the girl + sled immediately after their collision. Use momentum conservation to find the girl's speed just before she collides with the sled. Use energy to find the girl's initial height h.

Explanation / Answer

a)f = N = 0.12 x (40 +10) x 9.8 =  58.8 N

energy = work done by friction = f.d = 58.8 x15 = 882 J

(40 +10) v^2/2 = 882

v = 5.94 m/s

b) (40 +10) x 5.94 = 40 x u + 10 x 0

u = 7.42 m/s

c) mv^2/2 = mgh

7.42^2 = 9.8 x h

h = 5.625 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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