Question

A mass m = 12 kg is pulled along a horizontal floor with NO friction for a distance d =6.5 m. Then the mass is pulled up an incline that makes an angle θ = 25° with the horizontal and has a coefficient of kinetic friction μ_k = 0.44. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 25° (thus on the incline it is parallel to the surface) and has a tension T =45 N.

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)J

2)What is the speed of the block right before it begins to travel up the incline?m/s

3)What is the work done by friction after the block has traveled a distance x = 2.7 m up the incline? (Where x is measured along the incline.)J

4)What is the work done by gravity after the block has traveled a distance x = 2.7 m up the incline? (Where x is measured along the incline.)J

How far up the incline does the block travel before coming to rest? (Measured along the incline.)m

Answer 1

QM GTsinas t=450 uks 0.44 m=12kg l=0 ng T=45N XT=45N ezo (1) kd - = 6.5m WWT = TO CO325 =(45)(6.5)(0.90) W/7 = 263.25 de Joule u By work energy theorem WE AK LOT - Ale As WN, wing are zero - along Horgentel 963.25= Impro 2X263.25= 12 v² = v=1526.5 = y = 43.875 - V = 6.62mls 12 2m Wy=dad' cos180 = - diedl - Cuk Ndl whore N= ngens =(- Mking cos 25)dl = -0.44 X 1289.8 (0.90) 2.7 I g =-125.74 Joule te=MIN

1-2.1 work done by gravity 180° 900 Wg = (masinas) d'Cos 180 W by my cesas > masinasta nice25 ongcandst dl = = 12x9.8 (0•42) (:7) wg = - 133.36 Joule / whore V = 6.62 m/ let d he total distance on incline plone on which block more before comiy to Rest By work energy theorem Wg twop tot = sk -ngsin25d" +Td"-(MKN)d" = 0-1 mv² e d ing F- tal (ong sinas-T t'liking as 25) = Amy² d my 2 (ingsin 25-T + uking costs) - 12 ( 6 6 2 ) my? 1928.gan (1280.8 X (0.42) - 45+0.44X1267.880.90) 525.8928 (49.392 - 45 +46.5696 24/7 1888.874 Next Page

525. 89 28 2x 50.9616 - 525.8928 101.9232 d" - 5.159696 lally 5.16m