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PROBLEM 3 - 8 points [2 points per part] Pat the Pirate is digging up some of hi

ID: 1785140 • Letter: P

Question

PROBLEM 3 - 8 points [2 points per part] Pat the Pirate is digging up some of his buried treasure, in a cave in the hills. The treasure chest is heavy, but fortunately there is a rail car from a mining operation nearby, and Pat loads the treasure chest into the car and jumps in. Off he goes, down the track toward the ocean below. The combined mass of Pat, the treasure chest, and the rail car is 400 kg. Use g-10 N/kg. (a) Starting from rest, Pat's rail car drops through a height of 45 m. Neglecting all resistive forces what is the speed of the rail car at this point (the bottom of the hill)? (b) The bottom of the hill is shaped like a vertical circular arc with a radius of 150 m. Pat feels heavier than usual because of this - what percent heavier? (c) At the bottom of the hill, there is an empty rail car, with a mass of 200 kg, at rest on the track. Pat's rail car collides with this one, and they couple together, moving as one after the collision. What is the speed of the combined system after the collision? (d) Between the bottom of the hill and the ocean, the track goes 5 m vertically up a small hill. Because of a sticky brake on the empty rail car, the two rail cars barely make it over this hill before bringing Pat down to his rowboat on the shore. Between the point of the collision and the top of the small hil, how much work do resistive forces do on the two-car system?

Explanation / Answer

(A) Applying energy conservation,

PEi + KEi = PEf + KEf

m g h + 0 = 0 + m v^2 /2

v = sqrt(2 x 10 x 45) = 30 m/s

(B) N - m g = m v^2 / R

N = m (10 + 30^2/150) = 16m

N0 = m g = 10 m

% heavier = (16m - 10m) x 100 / 10m

= 60%

(C) Applying momentum conservation,

400 x 30 = (400 + 200) v

v =20 m/s

(d) Applying work - energy theorem,

Work done by gravity + work done by force = changein KE

(600 x 10 x - 5 ) + W = 0- (600 x 20^2 / 2)

W = - 90000 J

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