Question

A block of mass m= 7.30 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Acum , -7.50m) and (a) Determine the block's speed at points point point m/s to point (b) Determine the network done by the gravitational force on the block as it moves from point

Answer 1

given (M=7.3 kg [VA =o) ha = 7.5 m) 3. 2m / Apply energy conservation between point A and point ® (PE) + (KE) = (PE) B + (KE)B mgha + 1 mv² = mghe + Imve? 7,389.8)x7.5 +4x7,3 *O*= 7,3 X9-81X3-2 +4x743 * Vg? 73.575 + 0 = 31.392 + VB 73.575-31.392 = vs 42.183 x2 = VB V² = 84.366 YB = V84.366 VB =9. 185 m/s point C APRly energy conservation between point and (PEA + (KE) = (PE) c + (KE). ingha + 1 m V² = mghe + Įmų?

743x9-9187.5 +4x7/3X03* = 7,389.8IX 2 + 4,7,3 % Vě 73.575 to = 19.62 + 1 ve ² 73.575-19.62 = ve? 53.955 x2 = V? Vez = 107.91 Ve=1107.91 Ve = 10.387 m/s I Point B VB =9. 185 m/s Aus point @ Ve=10.387 m/s 1 - Ans on block from Net work point done by to point gravitional force t is W AC WAC= mg x Cha-ha) Wac = 7.3 x9.81 * (7.5 -2) Wac = 7.3 X9.81 X 5.5 TWAC=393.87 I Ans