Question

Two blocks m1=8.1 kg and mass m2 are connected by a massless cord over a massless pulley as shown below. the block of mass m2 is placed on a rough inclined surface at an angle (theta = 55)

Two blocks my = 8.1 kg and mass m2 are connected by a massless cord over a massless pulley as shown below. The block of mass mz is placed on a rough inclined surface at an angle = 55°, and a force of F=132 N is applied on mass my in the perpendicular direction to the rough surface as shown in the figure below. If the mass my is moving with a constant speed down the incline, find the value of the mass mz in units of kg. Take the coefficient of kinetic friction between the blocks and the surface as *-0.49 and g=9.8 m/s2. Round off your result to 1 decimal place. HA

Answer 1

So, friction force on block mi is fr = M, N, = Mx (Fitmig) friction force on block me is and f = MkN2 Mx m, gloso Now, as my is moving Therefore, net force acting on me with constant speed on me is hero س( Hence, we can write, mag sino - fa-fi = 0

5 %, 9 Sims — My m.goco — м (+, + mya) = 0 ү) Эm, (sing – A Clas B) = а( 4 m2 Mu (h+mig) g (sino ми Cos 0) С. 40 (192 + %- 1 X-8) 9-% (sin 56' - о - 40) Ces 551) 1 m 2. ka m) - - 10.64 ka The refore, the value of mass, m) = 10 - 64 и - - - с