Lesson 3 - 11)a) F smax = F g sin 45 o F smax = mg sin 45 o...

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

View Full Document Right Arrow Icon
Lesson 3 9)a) b) Static friction is the force that causes the small box to accelerate horizontally. c) F(F)=F(A) u s F(N) = ma u s mg = ma u s = ma / mg u s = a / g u s =(2.5 m/s 2 ) / (9.8 m/s 2 ) u s = 0.26 The coefficient of friction is 0.26 10) F a = F g – F f F a = F g sin5 o - u k F n F a = mg sin5 o – u k mg cos 5 = -27 N F = ma a = (-27 N) / (25 kg) a = -1.1 m/s 2
Background image of page 1

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

View Full DocumentRight Arrow Icon
t = (v f – v i ) / a = (0 - 3.5m/s) / -1.1 m/s 2 = 3.2s It takes the skier 3.2 seconds to come to a stop. b) d = v i (t) + 0.5(a)(t) 2 d = (3.5 m/s)(3.2s) + 0.5(-1. 1m/s 2 )(3.2s) 2 d = 11.2 m - 5.632 m d = 5.6 m The distance the skier travelled down the hill was 5.6 m.
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 11)a) F smax = F g sin 45 o F smax = mg sin 45 o F smax = (22 kg)(9.8 N/kg) sin 45 F smax = 152.5 N The maximum force applied upward, parallel to the ramp is 1.5 x 10 2 N. b) F n = (F smax ) / (u s ) F n = (152.5 N) / (0.78) F n = 195.5 N F a = F n F g cos45 o F n = 195.5 N (22 kg)(9.8 m/s 2 ) cos45 o F a = 195.5 N 152.5 N F a = 43 N The smallest force that can be applied to the top of the box is 43 N....
View Full Document

This note was uploaded on 12/13/2009 for the course ENGG eg 404 taught by Professor Weisenhaupt during the Spring '09 term at University of the Bío-Bío.

Page1 / 2

Lesson 3 - 11)a) F smax = F g sin 45 o F smax = mg sin 45 o...

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