212samplefinalsoln - 212 Sample Final exam Solution Dec,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
212 Sample Final exam Solution Dec, 2004 1) A truck applies a torque to its wheels resulting in a forward force on the truck of 500 pounds. The truck itself (without the crate) has a mass of 70 slugs. A 30 slug crate sits in the back of the truck on a horizontal surface with coefficient of friction μ = 0.1. As the truck accelerates from rest, the crate slides backwards relative to the truck. The free body diagrams for this system are: a) Solve for f, and for a crate and a truck relative to the ground If the crate starts a distance of 10 feet from the back of the truck, b) How long will it be before the crate collides with the back wall? c) What speed will the truck be going right before the collision? d) What velocity (speed and direction!) will the crate have relative to the ground right before collision? e) If the crate sticks to the back wall, what speed will the composite body have right after the collision? Neglect the rotary inertia of the wheels. a) f = m crate a crate , but f = μ N = μ m c g, so a crate = μ g = 3.22 ft/sec 2 to the right! 500 - f = m truck a truck , so a truck = 5.76ft/sec^2 to the right b) -10 feet = (1/2) a rel t 2 , with a rel = -2.54 , i.e., 2.54 to the left. so t = 2.81 seconds c) a truck x 2.81sec = 16.19 ft/sec to the right d) a crate x 2.81 = 9.05 ft/sec to the right e) m T 16.19 + m c 9.05 = L total = 1404 slug ft / sec NB this is also 500 lbf times 2.81 seconds ! = (m T +m c ) v final . So v final = 14.04 ft/second 2) An ice skater, of mass 70 kg, going a speed of 6 m/sec, collides with another ice skater (mass 50 kg) who is initially at rest. They entangle and fall to the ice, and then slide with a coefficient of friction of 0.12 How much time elapses until the composite comes to rest? What distance do they slide? Analyze collision: 70 kg * 6m/sec + 50 kg * 0m/sec = 420 kg m/sec = L initial = L final = 120 kg * vf so vf = 420 / 120 = 3.5 m/sec = speed of composite after collision Analyze sliding using impulse momentum: L initial = 420 kg m/sec = Lfinal(=0) - f t, where f = μ Mg = (.12) (120) (9.81) = 141.3 so t = 420/141.3= 2.97 seconds
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

Page1 / 4

212samplefinalsoln - 212 Sample Final exam Solution Dec,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online