An 8.70-kg block slides with an initial speed of 1.56m/s up a ramp inclined at a

Question

An 8.70-kg block slides with an initial speed of 1.56m/s up a ramp inclined at an angle of 27.4 degrees with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.62. Use energy conservation to find the distance the block slides before coming to rest.

Explanation / Answer

..........N................C ..........*................./ ................*........./ ......................*./A ....................../...|.....D (mg cos26.4) .(mg Sin26.4) /...mg ................B./__26.4 degrees__ Consider a (BAD) diagram as I could draw to make you understand better Let the block of mass m = 8.7 kg is on the ramp (which is at 26.4 degrees from the ground, at point A, so weight of the block will act vertically downwards and will have two components in the direction AD as mg cos 26.4 and towards B as mg sin 26.4 Since friction acts against the motion, which is uphill so friction will acts towards down side along with mg sin 26.4 in the direction of AB and is given as mu*R (mu = 0.52, R = mg cos 26.4) so Total resistance force acting downwards = mg sin 26.4 + 0.52*mg cos 26.4 where "m" is the mass of the block in kg resistance force = 8.7*9.81*sin 26.4 + 0.52*8.7*9.81 cos 26.4 = 77.7 N since F = ma ; where a = retardation a = F/m = 77.7/8.7 = 8.931 m/sec^2 since we know final vel is zero so v^2 = u^2 - 2as ;u = 1.56 m/s, s = distance, v = 0, a = 8.931 0 = 1.56^2 - 2*8.931*s s = 0.136 m ------------- Answer --------------------------------------… FROM the conservation of energy : KE = PE + work done against friction 1/2 mv^2 = mgh + F*d F = mu* mg cos 26.4 =>0.52*8.7*9.81*cos 26.4 = 39.75 N, d = distance on ramp, h = vertical height 1/2*8.7*(1.56)^2 = 8.7*9.81*{d*sin 26.4} + 39.75*d ; sin 26.4 = h/d so h = d* sin 26.4 10.5861 = 77.6982 d or d = 0.136 meter So both way you get the same answer, so it will slide 0.136 meter before it stops Hope this helps Hope this helped

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