Question

A package of mass m = 2.00 kg is released from rest at the top of an inclined plane as seen in the figure. The package starts out at heighth = 0.200 m above the top of the table, the table height is H = 2.00 m, and 0 = 50.0° R- (a) What is the acceleration (in m/s) of the package while it slides down the incline? 7.50 m/s² (b) What is the speed (in m/s) of the package when it leaves the incline? m/s (c) At what horizontal distance (in m) from the end of the table will the package hit the ground? 03171 (d) How long ins) from when the package is released does it to the ground

Answer 1

RAN Sino a AB Jo mgling ay BC Arch mgcose | mg ACC problem, As, there we assume is no mention of inclined plane friction in the is friction dess. the from the free body diagram of the package, equating force along x axis and y.axis , we get . mgline = ma 8 mg eso - N (normal reaction) :a = g Sing -- Acceleration of the package, a =9.81 x Sin 50° mys? = 17.51 mys2 AB Let the velocity of package at C = 0, when it Slides down from B et rest. So, the package's displacement = BC . Se mere 09 гд, ou u?=u²+zas 02 20 +2.gļine. Frience = 29. AB = 2gh => 0 = N2gh ----(2) EN2x9.81 x 0.2 mys = 1.98 m/s Time takes to travel o to a Ft. ovo utat => ť = = yols = 0.265

royale su cosa بہ This is a brojectile motion, of initial velocity U21.98 mys. As gravitational acceleration acts doronward, . horizontal component of v=vest is constant. Along downward vertical direction, initial velocity, u= usino = 1.98x Sin 50° mis = 1.52 mys acceleration = g = 9.81 mys2 and, distance travelled along vertical s= H=2m - s = ut + hat -- (3) → 2 = 1.52 + + x 9.81x42 = 4.90+2 +1.527 -220 ta -1.92 + N (1.59) + 484.9x2 2x4.9 1 t = 0.50 or t = -0.81, Time can not be negetine Along horizontal direction Rau cosoit = 1.98 x Corsoºx 0.5 (9 B =[0.64 m Total time = t + t = (0.26 +0.5075 = 0.765 1 e) The above answers are obtained using equili), Ran12) ear(s) and equilu), which are independent of mass of the package. No. Package mass does not effect.