Question
Physics (8th ed.) by Cutnell, Johnson Chapter 4, Problem 14 When a parachute opens, the air exerts a large drag force onit. This upward force is initially greater thean the weight of theskydiver and, thus, slows him down. Suppose the weight of theskydiver is 915N and the drag force has a magnitide of 1027N. Themass of the skydiver is 93.4 kg. What are the magnitude anddirection of acceleration? Cramster's solution is: Sum of forces = force of drag - weight So...F=1027N - 915N=112N As per Newton's 2nd law: F=ma So.. a=F/m a=112N/93.4kg=1.199m/s2 upward direction Why is it that cramster's answer ofa=112N/93.4kg=0.199m/s2 upward direction??? Thanks! Physics (8th ed.) by Cutnell, Johnson Chapter 4, Problem 14 When a parachute opens, the air exerts a large drag force onit. This upward force is initially greater thean the weight of theskydiver and, thus, slows him down. Suppose the weight of theskydiver is 915N and the drag force has a magnitide of 1027N. Themass of the skydiver is 93.4 kg. What are the magnitude anddirection of acceleration? Cramster's solution is: Sum of forces = force of drag - weight So...F=1027N - 915N=112N As per Newton's 2nd law: F=ma So.. a=F/m a=112N/93.4kg=1.199m/s2 upward direction Why is it that cramster's answer ofa=112N/93.4kg=0.199m/s2 upward direction??? Thanks! Why is it that cramster's answer ofa=112N/93.4kg=0.199m/s2 upward direction??? Thanks!
Explanation / Answer
net of forceF = force of drag - weight F=1027N - 915N=112N It act upwards As per Newton's 2nd law: F=ma So.. a=F/m a=112N/93.4kg=1.199m/s2 upward direction