Question

A 76.0 kg mass of is slid up a Q = 26.0° incline by a motor driven cable that runs parallel to the incline. It moves at a constant 1.5 m/s speed for a a distance of 55.0 m up the incline. The coefficient of kinetic friction between the mass and the incline is uk = .32. (a) Calculate the total work done by the motor to move the mass in kilojoules. X kj (b) Calculate the average power produced by the motor during this process. Note: 1 hp = 746 W hp

Answer 1

a) N acolconstant velocity) 0800 bus masina Fu = MAN ng 0=26 if, n is the contact normal force with the incline plane; theri the mg coso component of weight balanced by Z Will be So N=mo caso is acting The Kinetic friction, fx=MKNanking caso along with ng sino component of weight downward along the Plane. if is the force force given by the motora the object then, by ana law T on Newton's Ta framgsindama for Constant velocity, aso

= ) T= fkt mgsino 2) T=engcaso + pristu > 8802wtoris) buat ginen; the covered distance d = 55m m so, Worik done WE Td = mad (sino the caso) me = @ giver, m=76 kg 26° ura 0.32 d =.5sm e we know that ģ=9.8 m/s² w = 629.8355)-[sing + (0.32x (0526")] J W = 29739.25944J D བ; 29.73925944 KJ z ) 29.74 kJ / Cappron) W = n

the time tamen to cover distance d with speed ta & so, the average Power = ngu (sino + Macaso P- putting 76 kg, = 1:5 mis 0=26%, ur=0.32 g 9.8 mysl . 3 P-26-48*15) [Sin24't (0.32* (0326) 8110707119 w > 2 ) p 811:0707119 7A6 hp 1.087 226 155 hp U p=1.09 hp (appron)