Question

A block of mass m = 200 g is connected by a string of negligible mass to a second block of mass ma 400 g. The blocks are at rest on a rough table with coefficients of kinetic friction of = 0.300, and 2. H20.200. A force of 2.50 N in the positive x-direction is applied to mass m. mi m2 Find (a). the acceleration of each block. [5 marks] (b). the tension in the connecting string. [5 marks) The system slides down with m, in lead an inclined plane while attached by the string parallel to the plane. The angle of incline is 0 30. The coefficient of kinetic frictions between blocks and the incline are same as earlier. Compute the acceleration of the two boxes. [5 marks (c). (d). the tension in the string. [5 marks] (e) How would the answers to (a) and (b) change if m, trailed m2? [5 marks]

Answer 1

(2) М, 29 а = 0.9 km — Чао је • Ца Дk = 0-3 мкго - 2 LE=2.5 N E-T – A (m, a) - M, а -- — Л. (м, 2) = м, а — Е - це, (м 2 ) – че (м3) = 6, +мJQ Е- (-3)(0:4x10) — ф. 3(o-2xја) – (046) а E— 0-3 — . 6 = 0-6а • С — 4 Чао- 6 = . . . а , = = 1 - 83м) т— (0-1)(o-Axio) = (-а) () 1. 1 = [o1 466 л бг 6 (6) . Т. — (1) T

AZ Mog Singaa - Mg grus May N = Mgaus 30 264 (10) (13) = 213 Newton mgau 30 - 3 Newton T a lk a = (0-2)(263) = 0.483 The UK, M = (0-3) (33) = 0.353 on block me force - Mog sin so = (0.4(LO) (%) = aN M; g sim so = (0.2) (10)(1) = IN We see that component of weigh outweight the force of frictions So blocles, will more. with Common acceleration let us say á Magn30 - T-M N = m₂ a. Hug--014530406 IL

Mga Singoº 46, Ny+T = M, le -0-313+7= Deda -044803+T 30.da : 10307_T = 0.4 a (0.6) a = 107874: I Ta = 22979 w/84 - t = 0.1154 n ) D e ho we can see that acceleratim of m is greater than m, if there is no stone between them in previous setup so hy ettergi Interchanging the mauses Result ini : o Tension is store become zeno ② m, will yoush in a & sliche down with Custant Common acceleration) (m + m₂) g Sinzo - MN, MEN = Mama) a - 3 - 0.453-0.313 0.6l.a . [a= 2:979 m/s2 i
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