Question

A box, mass m, slides without friction down an incline plane at an angle θ with the horizontal. The incline plane is free to slide without friction across the floor it is resting on. Find the vector acceleration of both the plane and the box with respect to the floor. Confirm your answer using both Conservation of Energy and Newton’s Laws.

Answer 1

Iso": tab att -> let a be the acceleration of incline plane (wedge) - with respect to the ground in left side. and b be the acceleration of box with respect to wedge (incline plane) .direction shown in sigure. f. B, D of wedge N=mgroso casas we am acol 9 и ---- mig >b sino

U o u take system Both wedge + Box in this system No external force is - present in the Horizontal direction . and Hence, according to Newton first law. a body remains in own state without any external force. Hence, Ef=0 het / force in ( Horizontal direction is zero. Ma + m (a- 60080) = 0 o Ma + ma - mbcoso = 0 - 0 a von> bcoso uu net ox (1) PONIIIIIIIINNNNNNNNNIILLU201011 li Tacoro a in macosor Umgsino LL LLL LLL Now second write down the equation of force in barallel to incline plane . mgsino = m(b-acoso) — tim gsino = b-acoso . b = gsino + acoso AN put the value of b in equation 6 Ma + ma - m (gsino +acoro) COX0 = 0 Ma + ma - mgsinocojo_ ma coi'o = 0 a (M+m_mcop²o) = mg sinocoso (a= mg sino colo M+ma mco8²0 .. .. .. . . .. ..

put the value of a in equation (1) b=g sino ta coso b= g sino + mg sino coso , como (M+m-m costa) b = gsino + mg sino corso Mum-mcosao b=gsinori + m costo Mtm-mcos²o b=gsinor mtm-mcos²o +mco. Mtm-mcos²0 b=gsinor Mam 7=gsinor mim Mam (1-cas²o1 Mamsin ol > → Now, we assume initially that acceleration of wedge with respect to Ground is a a = mg sino coso = mg sino Coso M+m(1-cos'o Momsinto ' + and we assumed acceleration of Block (bor) with res bect to the inclined plane (wedge). Hence dow= abi awon and 46 46 48 - ๒๕๒ a8G= a Bw tawo acceleration of wedge =ő ta respect = g sino mam 7+ mg sino cose to [mam since I Ma msinto Ground / with

Mam tmcoso . M+ msinto