Question

A 50 kg box is thrown along an inclined plane with an angle of 20, the initial speed of the box is 30 m/s. We know µ_s = 0.25 and µ_k = 0.20 between the box and the inclined plane find:

A) How much time does it take for the box to go up the inclined plane before coming to a stop?

B) The distance traveled by the box when going up the inclined plane

C) How much time does it take for the box to return to the starting point from which it was launched?

D) With what velocity does it return to the starting point?

50kg ) 20°

Answer 1

given v= 30m/sec Ms=0:25 Mk = 0.2 mass m=50kg Sokg 7200 au Walang a Free Body Diagram of mass NE Normal Reachon force on mass by inclined Plane mgsino mg mgcono 1 Equating force (parallel to inclined plane) a = acceleration en mg sino + fk = ma downword direchon equating force perpendicular fo= kmetic friction force to en clined plane 1. = MIN Nomgcopo fk = Mk N = Mk mg coso mas mgsino + MK mg coso a= géno + MK COOO) - accelasahon of Block in downword direction

In dioection pcoallel to enclined blane at stop condition final Velocity Vf = 0 . Initially Velocity of mass Vi = V-30 m/sec Vf= Vi -at 0 =V23. - [9ino + MK 2000)] + 981not MK 2000) t = 30 - 9.81 [sin 20 +0.2 20120] My B It=5.7704575 sec/ Answer part ② =time to stop distance toavel by mass before stop = 8 Vf2= Vi2 – 298 - Vi 2_vg2 302-02 2x9.81[11020 +0.200320] TD= 86.556862 m) Answer past® 2 a velocity at top point reache by mass to Mizo VRizom/sec

FBD of mass during Return velocity ) .Voico (Return inehal Equating force berpendiceelar to inclined plane Nemg cuno a mg sino 1. Doro mg0080 maximum value of Static friction fore fs) max = MS N = 0.25 Xmg coso = 0-25X50X9.818COD20 net clown word froce in statec condition = mg sino - (fs) max fef Negative - 1 then mass not E U39.81X81120850 goes in clownwood -0.25X50X9.81XCOD20 clirector} = 52.53107268 BN hence mass moves en down word direction fa kinetic faction force => fx = MAN f = fx = MK mg cono hence , AR= Retuon accelarahon mar = mgsmo-f mar = mg sino - MK mg coso ар = a km 0 — Hк соғ0)

&= Vai tp + 1/2 AR dp ² 8 h = oxfRt 1 x g(sino-MK 2000) AR² AR= | 25 J 968100-MR COBOJ fez / 2* 86-556862 TR v 9.81€ sin 20 -0.2% Cox 20°) te = 10.761575 see = 10.70177532 sec Answer part © Return time to reach initial position veloceity cyteo Rutuim sto initial position avef VR f = Vritate Vef = 0+ g(3ind – MK Coro) tr. = 9.81 [ sin 20 -0.260x20] x 10-70177532 NRS = 16e176168 m/sec) Answer parto