Chapter 7 Energy and Work Alternative method for the study of motion In many ways easier gives additional information Kinetic energy consider an object of mass m and speed v we define the kinetic energy as 2 1 K 2 a scalar not a vector units kg m2 s2 N m Joule J in S I ft lb in British units and erg in CGS like speed gives a measure of an object s motion a car and tractor trailer may have the same v but different K E mv K Work the work done on an object by an applied constant net force F which results in the object undergoing a displacement of d or r W Fd d F cos d F d cos a scalar units of N m J if F and d are perpendicular W 0 work can be negative 90 F Fd d Work Energy Theorem when a net external force does work on an object there is a change in the object s KE W W total K K f K i 2 f Wtotal mv mv 1 2 1 2 2 i Example A crate on a incline is held in place by a rope The rope is released and the crate slides to the bottom Determine the total work done if the crate has a mass of 100 kg the incline has angle of 50 0 the coefficient of kinetic friction is 0 500 and displacement of the crate is 10 0 m Solution Given m 100 kg 50 k 0 500 d 10 0 m Approach compute the work for each force n fk y n fk d x mg mg FBD d Only force components along the direction of d contribute x direction F y N mgcos 0 N mgcos W N N cos90 d 0 W f F cos d f k cos180 d f k d k mgcos d 3 15x10 3 J W g mgcos 90 d mgsin d 3 7 51x10 J Total work W total W g W f mgsin d k mgcos d mgd sin k cos 7 51x10 3 3 15x10 3 J 4 36x10 3 J Or calculate the net force along d x direction F x mgsin f k mgsin k mgcos mg sin k cos Fd max W Fd d mgd sin k cos Same as above Now determine final velocity from work energy theorem since vi 0 Ki 0 Wtotal K f K i mv 1 2 2 f 2W 2 4360 J vf 9 34 ms m 100 kg Check by kinematics 2 f 2 i v v 2ax x v f 2ax x 2g sin k cos x v f 9 34 ms Example A hockey puck slides across the ice Its speed slows from 45 00 m s to 44 67 m s after traveling a distance of 16 0 m Determine the coefficient of kinetic friction between the ice and the puck Solution Given vi 45 00 m s vf 44 67 m s x 16 0 m d Method Use work energy theorem FN d Fy FN mg 0 FN mg fk mg F x f k max f k k FN k mg W F cos d f k cos180 d f k d k mgd W K f Ki 12 mv 2f 12 mv 2i k mgd v 2f v 2i 2 k gd 2 f 2 i 2 2 v v 44 67 45 00 k 2gd 2 9 80 16 0 0 094
View Full Document
Unlocking...