Physics > QUESTIONS & ANSWERS > University of California, Los AngelesPhysics 6CHChpt7_ExampleProbs_wSols1b (All)
Problem 7-6 A particle of mass 0.500 kg is shot from P as shown in the figure. The particle has an initial velocity vi G with a horizontal component of 30.0 m/s. The particle rises to a maximum h... eight of 20.0 m above P. Using the law of conservation of energy, determine (a) the vertical component of vi G , (b) the work done by the gravitational force on the particle during its motion from P to B, and (c) the horizontal and the vertical components of the velocity vector when the particle reaches B.Problem 7-6 solution = 0 vy 0 ground level unknown component not !) (only given horizontal 30 m/s 0.5 kg = θ = = m V y i ix Kv T 20.0 h a) Determine viy. Use conservation of energy. We need 2 points, we select points P and T. If we call P initial and T final we get 2 ( ) 2 12 12 E mv mg H h E mv mgH E K U f i i f i i i = + + = + = +v i v ix note that 2 2 2 Vi = Vix +Viy also 2 2 2 2 2 V f = V fx +V fy = V fx = Vix horizontal velocity doesn’t change for a projectile (Chpt 3)! 0 at max height So use Ei = E f v iy (could determine now if wanted) 12 12 12 2 19.8 m/s ( ) 2 2 2 2 = = θ = + + = + + V gh mV mgh m V V mgH mV mgH mgh iy iy ix iy ix (b) Work done by gravity from launch are P to B, so our 2 points are P and B (limits of integration) W = 0∫ F dy = −mg y 0H = −mg(0 − H ) = mgH = 294 J H g g(c) vG at point B horizontal and vertical components. Take the 2 points at P for initial and B for final. Recall that Vx doesn’t change, so Vix = Vx = 30 m/s horizontal component ( ) ( ) ( ) 30.0ˆ 39.6 ˆ m/s 2 or 2 39.6 m/s 0 2 2 2 2 2 2 2 2 2 12 12 12 12 i j V V gH V V gH mV mgH mV m V V mgH m V V y iy iy iy y ix iy x y y = − = + = ± + = + = + + = + + Gv must take negative signProblem 7-23 The coefficient of friction between the 3.00-kg block and the surface in the figure is 0.400. The system starts from rest. What is the speed of the 5.00-kg ball when it has fallen 1.50 m? [Show More]
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