Question

Question: A block is placed on an inclined plane as shown in Figure 1. The block has a mass m = 1.85 kg. The angle is slowly increased. When 0 = 39.5° the block starts to slide down the inclined plane. When the block has travelled 1.00 m along the slope it has a speed of v = 0.555 m/s. Figure 1: A block of mass m resting on an inclined plane angled at degrees to the horizontal. The plane slopes upwards from left to right. Part 1) What is the net force acting on the block as it starts to slide? net = Part 2) What is the coefficient of static friction between the block and the slope? Ms = 0.824 Your last answer was interpreted as follows: 0.824 Part 3) What is the coefficient of kinetic friction between the block and the slope? Mk = Check

Need help with the first and third question. Thanks

Answer 1

DATE: 1 1 PAGE NO.: m= 1.8 N = ng Caso mgsmo menye mg loso To mg No : 0 (rest) , v= 0.555 m/s d = 1.oom 12 Vo² = aad = 0.555²-0²= axax 1 =) a = 0.154 m/s² Feet = ma = 1.85 ×0.154 = (0.285N. at o= 39.5 f = masino =) lls N = mg sino Ms ng loso = masino = us - Sino a tano cose Ms = tan 39.5=10-824 Now Fret = masing-fkmg Sino - Mk N fret = masino - Mx ng caso le mg Loso = masino - fret ex x 1.8589.8 Cos 39.5 = 1.8589.8 x Sin 39.5 -0.285 elx = 10.804 |