solution_pdf

solution_pdf - ferri(lf7249 – HW6 – mackie –(10611 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: ferri (lf7249) – HW6 – mackie – (10611) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 3 . 6 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 19 . 4 N. The coefficient of kinetic friction is 0 . 139. The acceleration of gravity is 9 . 8 m / s 2 . Find the speed of the block after it has moved 3 . 21 m. Correct answer: 5 . 08442 m / s. Explanation: We have to use the equation Δ K =- f s (1) , to calculate the change in the kinetic energy, Δ K . The net force exerted on the block is the sum of the applied 19 . 4 N force and the frictional force. Since the frictional force is in the direction opposite the displacement, it must be subtracted. The magnitude of the frictional force is f = μ N = μmg . Therefore the net force acting on the block is F net = F- μmg = 19 . 4 N- (0 . 139) (3 . 6 kg) (9 . 8 m / s 2 ) = 14 . 4961 N . Multiplying this constant force by the dis- placement, and using equation (1), we obtain Δ K = F net s = (14 . 4961 N) (3 . 21 m) = 46 . 5324 J = 1 2 mv 2 , since the initial velocity is zero. Therefore, v f = radicalbigg 2 Δ K m = radicalBigg 2 (46 . 5324 J) 3 . 6 kg = 5 . 08442 m / s . 002 10.0 points You leave your 50 W portable color TV on for 7 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0 . 1 / kW · h, what would be your monthly (30 day) bill? Correct answer: 1 . 05 dollars. Explanation: Let : P = 50 W and t = 7 h / day , The energy consumed in each day is W = P t = (50 W) (7 h / day) · kW 1000 W = 0 . 35 kW · h / day . In 30 days, you would use (30 day) (0 . 35 kW · h / day) = 10 . 5 kW · h , which would cost you (10 . 5 kW · h) ($0 . 1 / kW · h) = $1 . 05 . 003 10.0 points A car of weight 3100 N operating at a rate of 207 kW develops a maximum speed of 41 m / s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e. , if θ is the angle of the incline with the horizontal, sin θ = 1 / 20 ? Correct answer: 39 . 7788 m / s. Explanation: ferri (lf7249) – HW6 – mackie – (10611) 2 If f is the resisting force on a horizontal road, then the power P is P = f v horizontal , so that f = P v h = (2 . 07 × 10 5 W) (41 m / s) = 5048 . 78 N . On the incline, the resisting force is F = f + mg sin θ = f + W 20 = P v h + W 20 . And, F v = P , so v = P F = P P v h + W 20 = (2 . 07 × 10 5 W) (2 . 07 × 10 5 W) (41 m / s) + (3100 N) 20 = 39 . 7788 m / s ....
View Full Document

This document was uploaded on 10/22/2010.

Page1 / 8

solution_pdf - ferri(lf7249 – HW6 – mackie –(10611 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

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