aguilera (ssa2289)-Homework #7-conservation 010 (part 1 of 2) 10.0 points A roll

Question

aguilera (ssa2289)-Homework #7-conservation 010 (part 1 of 2) 10.0 points A roller coaster cart of mass rn = 202 kg starts stationary at point A, where hi = 36.1 m and frict the and a while later is at B, where h2 = 13.2 m. The acceleration of gravity is 9.8 m/s2 . it What is the potential energy of the cart relative to the ground at A? m Answer in units of J. 011 (part 2 of 2) 10.0 points What is the speed of the cart at B, ignoring the effect of friction? ary 16,084 012 (nart 1 af 2 100 oints Answer in units of m/s.

Explanation / Answer

Part 2 ) We can use the energy conservation
The total energy of the system remains constant. The total energy of the cart at the position A is the potential energy at the point A
E = m g h = 202 kg x 9.8 m/s2 x 36.1 m
E = 71463.56 J
The total energy at the position B is the sum of kinetic and potential energies. The potential energy at the position B is
PE = m g h = 202 kg x 9.8 x 13.2
PE =  26130.72 J
Since the energy is conserved, the total energy at the position B is equal to the total energy at the position A
EB = PE + KE = EA
The kinetic energy at B
KE = EA - PE
KE = 71463.56 J - 26130.72 J
KE =  45332.84 J
(1/2) m v2 = KE
v = sqrt (2 KE / m)
v = sqrt (2 x 45332.84 / 202)
v = 21.19 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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