Data for Honors Physics Chapter 5 Created for CVCA Physics By Dick Heckathorn 21 Nov 2K + 3

Table of Contents 3 Acceleration, latitude, radius 4 Planet, accsurface, ratio ap/ae 5 Assumptions about Strings & Pulleys 9 Page 165 #2 12 Page 165 #3 16 Coefficient of Friction

Latitude g Distance to (N/kg) Center (km) 0 9.7805 6378 15 9.7839 6377 30 9.7934 6373 45 9.8063 6367 60 9.8192 6362 75 9.8287 6358 90 9.8322 6357

Planet g or ag gp/ge Mercury 3.52 0.36 Venus 8.52 .87 Earth 3.72 .38 Mars 9.81 1.00 Jupiter 26.10 2.66 Saturn 11.07 1.12 Uranus 10.49 1.07 Neptune 13.80 1.40 Pluto 0.31 0.032 Moon 1.61 0.16

Assumptions about Strings & Pulleys - 1 Strings are considered to have negligible mass and are capable of exerting only “pulling” forces on objects to which they are attached (tension forces).

Assumptions about Strings & Pulleys - 2 Strings transmit forces undiminished: the tension force in a string is the same throughout its length.

Assumptions about Strings & Pulleys - 3 A frictionless pulley changes the direction of a string without diminishing its tension.

Assumptions about Strings & Pulleys - 4 Strings are assumed not to stretch.

Problem Page 165 #2 A 40 kg block on a level, frictionless table is connected to a 15 kg mass by a rope passing over a frictionless pulley. What will be the acceleration of the mass?

Problem Page 165 #2 Ft on b Fe on b Fe on b 147 N Fnet a = = 147 N = 15 kg x 9.8 m/s2 2.7 m/s2 55 kg

What is the force on 40 kg mass? Where is the rest of the 147 N? What is the tension on the string? What is the force on 40 kg mass? 107 N Fon 40kg = 40 kg . 2.67 m/s2 107 N m . a F = 147 N a = 2.7 m/s2 Fon 15kg = 15 kg . 2.67 m/s2 m . a 40 N

Problem Page 165 #3 A 3.0 kg mass is attached to a 5.0 kg mass by a string that passes over a frictionless pulley. When the masses are allowed to hang freely, what will be the acceleration of the masses and the magnitude of the tension in the string?

Problem Page 165 #3 Fe on b Fe on b Fnet 19.6 N Fnet a = 2.45 m/s2 = (5 – 3) kg x 9.8 m/s2 = 19.6 N 8 kg

Problem Page 165 #3 Fa Fs on b Fe on b Fstring = T = Fs on b + Fa T= 3.0 kg x 9.8 m/s2 +3.0 kg x 2.45 m/s2 T = 29.4 N + 7.35 N = 36.8 N

Problem Page 165 #3 Fa = 7.4 N 36.8 N 29.4 N 29.4 N Fa = 12.2 N 36.8 N 61.2 N 49.0 N

Surface μ oak on oak, dry 0.30 Waxed hickory-dry snow 0.18 Steel on steel – dry 0.41 Steel on steel – lubricated 0.12 Steel on ice 0.01 Rubber on asphalt, dry 1.07 Rubber on asphalt, wet 0.95 Rubber on concrete, dry 1.02 Rubber on concrete, wet 0.97 Rubber on ice 0.005