Question

2- A block of mass is on an inclined with a friction coefficient equal to over a massless, frictionless pulley to a second block of mass 0.1 it is connected by a cord MA 320g 09 300 • Draw the free-body diagram and write Newton's second law equation for each block. • Calculate the tension in the cord. • If block B start from rest, calculate the displacement of after 1 s. • Calculate how much mass we need to add to the block B that will make moving it at a constant velocity.

Answer 1

Ins. (a) Free m9 free body diagram is a m-032 kg 0-30° my = 0.87 kg, 45ool - maglio o od Ff het the block 2 moves down ward with toimi m o acceleration 'a'. (6) Applying Newton's law on ma 2 a FR = MR m. g - I = m₂ a on , T-m, g sino p m glasu = ma - mg (g-a) - m glino - um, g650 am, a a = magam, glino-pm, glose m, tina a = 5.68 m/s2 Tension - mg 19-9) - 3.64 N s= ut tlat? to displacement after le 5.62 w2 = szot 2.81 m Td For con relaci . constant acceleration =0 a = mag - mg sino - M m, gloso =0 . m, tma - ma - msing + Mn, Coro = 0.19 kg - So, we need to take off (0.87-0.19) = 0.68 kg from mo Scanned with CamScanſfer