Question

An object released from rest slides down a frictionless ramp that makes an angle 0 of 70.00 with the horizontal. The object begins at a height H of 11.0 m above the base of the ramp. The bottom end of the ramp merges smoothly with a second frictionless ramp that rises at angle 02 of 35.0°. What distance d does the object slide along the second ramp before coming to a momentary stop? H e TH d= m When the object is on its way back down the second ramp, what is its speed up at the moment that it is at a height H2 of 5.00 m above the base of the ramp? Uf = m/s

Answer 1

Hi H2 o 2 2 1472 211 m -> h H, = 11 m, H₂ = 5m, 0,= 70°, O₂ = 35° Let m be the mass of the object. Potential energy of the object at height Hiis, Ep = mgh It visite speed of the object on waching the base of the vamp, it's kinetic energy, Ex = I mv. Applying Law of conservation of energy, Ex= Ep = I mv² mgH, = V2 = 29H, = 2x9.8 x 11 = 215.6 = v2 = 215-6 = 1214.7 m/s Let the object rises to a vertical height h on the second samp. = mgh, = ļ mv² h =v² 29 If d is the distance the object travels along the second sump, Sin O₂ = h h 19.2m d Sino, distance, d=19.2m is the speed of the object at a height Hy above the base, Ź my? 2 mg ch - H₂) = v= 29Ch - H₂) = 2x9.8 x (115) up 2 = 117.6 - Vf = 10.8 m/s = 10.8 m/s 2x9.8 > da 11 Sim 35 If Vf VA