c. Suppose the coefficent of kinetic friction was equal to 0.5. how much would t

Question

c. Suppose the coefficent of kinetic friction was equal to 0.5. how much would the thermal energy increase as the train travels along the horizontal surface?

d. How does your answer to part C compare to the inital total mechanical energy of the train? What does this imply about the motion of the train?

e. Using this larger coefficent of the friction, determine how far along the horizontal surface the train travels before coming to a stop.

struct mathematical equations of energy conservation using the basic energy model for a of variety of ations from mechanics dict forces on a system and motion of a system from graphs of energy oy velocity vo top of a hill. The train travels down the hill, then along a flat flat train has a non-zero at the and rizontal surface until collides an elastic bumper that brings the train to a stop. Assume the hill it with rface are frictionless and neglect air resistance. a. a schema L for this scenario. w system and choose an appropriate system

Explanation / Answer

b)

consider the motion from A to B

Vo = initial velocity at top = 20 cm /s = 0.20 m/s

Vf = final velocity at the beginning of the flat surface = ?

h = height = 15 cm = 0.15 m

using conservation of energy on the incline plane

(0.5) m Vo2 + mgh = (0.5) m Vf2

Vf2 = Vo2 + 2gh

Vf2 = (0.20)2 + 2 (9.8) (0.15)

Vf = 1.73 m/s

L2 = length of flat rough surface = 0.40 m

fk = frictional force on the flat surface = uk mg

Wk = work done by the frictional force

Vf' = velocity of train just before hitting the spring

using conservation of energy

(0.5) m Vf2 = Wk + (0.5) m Vf'2

(0.5) m Vf2 = uk mg L2 + (0.5) m Vf'2

(0.5) Vf2 = uk g L2 + (0.5) Vf'2

(0.5) (1.73)2 = (0.3) (9.8) (0.40) + (0.5) Vf'2

Vf' = 0.801 m/s

m = mass of train = 1.2 kg

k = spring constant = 90 N/cm = 9000 N/m

x = compression of the spring

using conservation of energy for train -spring collision

spring energy = kinetic energy of train just before collision

(0.5) k x2 = (0.5) m Vf'2

(9000) x2 = (1.2) (0.801)2

x = 0.00925 m

c )

increase in thermal energy = new work done by friction - old work done by friction

increase in thermal energy = u'k mg L2 - uk mg L2 = (0.5 - 0.3) (1.2) (9.8) (0.40) = 0.941 J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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