mid term 2 practice 2

# mid term 2 practice 2 - brown (twb493) – extra credit 04...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: brown (twb493) – extra credit 04 – turner – (57340) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points You’re driving a vehicle of mass 1350 kg and you need to go around a flat circular curve whose radius of curvature is 77 m. The coef- ficients of static and kinetic friction are both equal to 0 . 27. What is the fastest speed you can drive and still make it around the turn without skidding? Correct answer: 14 . 2738 m / s. Explanation: The equations of motion in the x and y directions are given by: summationdisplay vector F x = μ s N = mv 2 max r summationdisplay vector F y = N- mg = 0 . Note that the equation in the x direction uses μ s , the coefficient of static friction, since we are interested in motion without skidding. Furthermore, we use equality instead of ‘ ≤ ’ because we are interested in the maximum speed. Remember that static friction pro- vides whatever force is necessary up to the limit of F s = μ s N — once that limit is ex- ceeded, kinetic friction takes over. Using N = mg and solving the y equation for v max yields v max = √ μ s rg = 14 . 2738 m / s . 002 10.0 points What is the minimum speed that a roller coaster car of mass 510 kg must have in order to make it around the inside of a loop of radius 12 m without falling off the track? Correct answer: 10 . 8444 m / s. Explanation: The equation of motion in the y direction at the top of the loop is: summationdisplay vector F x = N + mg = mv 2 r To satisfy the condition of minimum speed without falling off the track, we let N → 0, so that the centripetal acceleration is provided entirely by the gravitational force. Solving the resulting equation for v yields v min = √ rg = 10 . 8444 m / s . 003 10.0 points A block is oscillating up and down at the end of a vertical spring. Complete the following statements: At the instant the block reaches its lowest position, the direction of the block’s momen- tum is I. up. II. down. III. not defined, since the magnitude is zero. At the instant the block reaches its lowest position, the direction of the rate of change of momentum dvectorp/dt is I. up. II. down. III. not defined, since the magnitude is zero. At the instant the block reaches its lowest position, the direction of the net force vector F net on the block is I. up....
View Full Document

## This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 6

mid term 2 practice 2 - brown (twb493) – extra credit 04...

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

View Full Document
Ask a homework question - tutors are online