1.. In Clss or Homework Eercise 1. An 18.0 kg bo is relesed on 33.0 o incline nd ccelertes t 0.300 m/s. Wht is the coeicient o riction? m 18.0kg 33.0? 0 0.300 m / s irst, we will ind the components o the orce o grvit: g mg sin (18.0)(9.80) sin 33.0 96.1 g mg cos (18.0)(9.80) cos 33.0 148 Perpendiculr orces (using w rom the rmp s positive) m m 0 148 g m g 148 Prllel orces (using down the rmp s positive) RRHS Phsics Pge 75
m (18.0)(0.300) 96.1 g g m m 90.7 (148) 90.7 0.613. A bo (mss is 455 g) is ling on hill which hs n inclintion o 15.0 o with the horizontl.. Ignoring riction, wht is the ccelertion o the bo down the hill? m 455g 0.455kg 15.0 0? irst, we will ind the components o the orce o grvit: g mg sin g mg cos (0.455)(9.80) sin15.0 (0.455)(9.80) cos15.0 1.15 4.31 Since there is no riction this time, we do not need to look t the perpendiculr orces: Prllel orces (using down the rmp s positive) RRHS Phsics Pge 76
m m m (0.455) 1.15 g g.53 m / s b. I there is coeicient o riction o 0.0, will the bo slide down the hill? I so, t wht ccelertion? m 455g 0.455kg g g 1.15 0.86 0.0 0? Since there is riction this time, we will need the norml orce. We must thereore look t the perpendiculr orces: Perpendiculr orces m m 0 4.31 g m g 4.31 Prllel orces (using down the rmp s positive) RRHS Phsics Pge 77
(0.0)(4.31) 0.86 Since g, the bo will ccelerte down the hill. m (0.455) 1.15 0.86 g g m m 0.64 m / s c. How much orce is required to push the bo up the rmp t constnt speed? Since the bo is now going up the hill, riction must be down the hill: m 0.455kg g 0.86 1.15 0 0 Using up the rmp s positive nd looking t the prllel orces, RRHS Phsics Pge 78
m m 0 1.15 0.86 g p m g p p g p g.01 3. A 165 kg pino is on 5.0 rmp. The coeicient o riction is 0.30. Jck is responsible or seeing tht nobod is killed b runw pino.. How much orce (nd in wht direction) must Jck eert so tht the pino descends t constnt speed? Initill, we do not know wht direction Jck hs to ppl his orce so tht the pino descends t constnt speed. I we look t the ree Bod Digrm without Jck s orce m 165kg 5.0 0.30 p 0 0? We see tht we must compre g nd g mg sin (165)(9.80) sin 5.0 683 g mg cos (165)(9.80) cos 5.0 1470 Perpendiculr orces: RRHS Phsics Pge 79
m 0 1470 g g m m 1470 (0.30)(1470) 440 Since g, the pino will ccelerte down the rmp i Jck does nothing; thereore Jck must push up the rmp: Using up the rmp s positive, Prllel orces: m m 0 p g p g p g 683 440 40 The orce must be directed up the rmp. b. How much orce (nd in wht direction) must Jck eert so tht the pino scends t constnt speed? RRHS Phsics Pge 80
g p 683 440 0? Using up the rmp s positive, m m 0 p g p g p g 683 440 110 The orce must be directed up the rmp. 4. A cr cn decelerte t -5.5 m/s when coming to rest on level rod. Wht would the decelertion be i the rod inclines 15 o uphill? The coeicient o riction will be the sme in ech sitution (sme surces). We must ind this rom the level surce prt o the problem. Level Rod 5.5 m / s? Using the orwrd direction s positive, RRHS Phsics Pge 81
m m m m mg 5.5 (9.80) 0.56 Incline 0.56 15 0? Since there is no perpendiculr ccelertion, g. Looking t the prllel orces nd using up the rmp s positive, m g g m mg sin mg cos m m (9.80) sin15 (0.56)(9.80) cos15 7.8 m/ s 5. A sled is sliding down hill nd is 5 m rom the bottom. It tkes 13.5 s or the sled to rech the bottom. I the sled s speed t this loction is 6.0 m/s nd the slope o the hill is 30.0, wht is the coeicient o riction between the hill nd the sled? Using down the rmp s positive, RRHS Phsics Pge 8
d 5m v 6.0 m / s i t 13.5s 30.0? d v t t 1 i 5 (6.0)(13.5) (13.5) 1.6 m / s 1 m m mg sin mg cos 1.6 (9.80) sin 30.0 (9.80) cos 30.0 0.39 g g g m m m 6. A 5.0 kg mss is on rmp tht is inclined t 30 o with the horizontl. A rope ttched to the 5.0 kg block goes up the rmp nd over pulle, where it is ttched to 4. kg block tht is hnging in mid ir. The coeicient o riction between the 5.0 kg block nd the rmp is 0.10. Wht is the ccelertion o this sstem? igure 19: Digrm or Question 6 RRHS Phsics Pge 83
m 5.0kg m 1 4.kg 0.10? We must irst determine the direction o ccelertion o the sstem b compring g nd g1. g1 m1g sin (5.0)(9.80) sin 30.0 4.5 g m g (4.)(9.80) 41. Since g g 1, the sstem will ccelerte clockwise nd riction on the 5.0 kg object will be down the rmp. I we tret the whole sstem s one object (s in Unit 3), our linerized ree bod digrm will look like this: The tension in the rope cn be ignored since we re treting the sstem s one big object. g mg cos (0.10)(5.0)(9.80) cos 30.0 4. Using clockwise s the positive direction (towrd the 4. kg mss), RRHS Phsics Pge 84
m m t g g1 m t g g1 (9.) 41. 4.5 4. 1.4 m / s 7. A orce o 500.0 pplied to rope held t 30.0 o bove the surce o rmp is required to pull bo weighing 1000.0 t constnt velocit up the plne. The rmp hs bse o 14.0 m nd length o 15.0 m. Wht is the coeicient o riction? igure 0: Digrm or Question 7 p 500.0 30.0 g 1000.0 0 0? g mg 1000.0 m(9.80) m 10kg RRHS Phsics Pge 85
We must brek the pplied orce up into components tht re prllel nd perpendiculr to the rmp. cos sin p p (500.0)(cos 30.0 ) (500.0)(sin 30.0 ) 433 50. p p et we must ind the ngle tht the rmp mkes with the horizontl so tht we cn ind the components o the orce o grvit: 14 cos 15 1 g mg sin (10)(9.80) sin 1 360 g mg cos (10)(9.80) cos 1 930 Perpendiculr orces m p g 0 p g 0 50 930 m 680 Prllel orces (using up the rmp s positive) m p g 0 0 433 360 m 73 p g 73 (680) 0.11 8. I bicclist (75 kg) cn cost down 5.6 o hill t sted speed o 7.0 km/h, how much orce must be pplied to climb the hill t the sme speed? We cn ssume the sme mgnitude or the orce o riction on the w up nd the w down the hill: Down the Hill RRHS Phsics Pge 86
m 75kg 5.6 0 v 7.0 km / h? Prllel orces (Using down the rmp s positive) m 0 g g m mg sin (75)(9.80) sin(5.6 ) 7 Up the Hill m 75kg 5.6 0 v 7.0 km / h p 7? RRHS Phsics Pge 87
Prllel orces (Using down the rmp s positive) m m 0 g p g p p g 7 7 144 RRHS Phsics Pge 88