I\'m so confused on how to work this problem. Please Help. By day, Dr. Kathleen
ID: 1828970 • Letter: I
Question
I'm so confused on how to work this problem. Please Help.
By day, Dr. Kathleen Jackson is a mild-mannered professor of engineering who enjoys teaching rigid-body dynamics. On the weekends and during the summer months she is a professional stuntwoman known as "Cannonball Kate" Jackson. In her latest stunt work, she portrays an investigative civil engineer who is being pursued by a group of employees working for the corrupt municipal water authority. The scene opens with Kate coming out of her third floor hotel room into the hallway to discover that the "bad guys" are walking toward her from both ends of the hallway. Quickly assessing the situation. Kate sees the open door of a laundry service room just across the hallway. Running into the room, she spies an open laundry chute for transporting laundry bags down to the first floor. Grabbing a plastic, room-service, food tray, she jumps on the tray and "rides" (kneeling) the tray down the chute to the first floor, (see schematic below) As she emerges from the chute on the first floor. Kate lands in a stationary, four-wheeled laundry cart that then rolls across the floor with no friction. The cart travels out of the first-floor laundry room, off of the edge of the loading dock and lands (hopefully) in the bed of a pickup truck driven by Kate's stunt partner. The question is, "does she make it into the pickup truck?" To answer this question, you need some more information. Kate weighs 105 lbs Kate has a zero velocity at the top of the chute (point A). Coefficient of friction between plastic tray and the metal laundry chute is 0.2 Neglect the weight of the plastic tray. Weight of the empty laundry cart is 100 lbs. Information you need to supply me. Kate's velocity at the bottom of the chute (point B)? The velocity of the laundry cart alter Kate lands in it (point B)? The speed of the laundry cart at the edge of the loading dock (point C)? Does Kate make it into the back of the pickup track (point D)?Explanation / Answer
1. For velocity at B use conservation of energy
At the top of chute, energy = mgh = 105*26
As she reaches the bottom the enrgy is converted into frictional energy loss and kinetic energy.
Frictional force= friction coefficient*Normal Force
Normal force = mg*cos45 = 105/sqrt(2)
Frictional Work = Frictional Force*distance travelled.
Distance travelled = 14/cos45 + 12/sin45 = 26*sqrt(2)
Let velocity at B = v
Mass of kate = mg/g = 105/32.17
105*26 = 105/32.17*v^2/2 + 0.2*105*26
v = 36.58 ft/s
2. After she hits the cart, we can apply Newton's second law in horizontal direction.
Since there is no impulse in horizontal direction, Momentum in horizontal direction is conserved.
mass of laundry cart = 100/32.17
Initial Momentumin horizontal direction = 105/32.17*36.58*cos45 + 0
Total Mass*Final Velocity = (100/32.17 + 105/32.17)*V
Since both are equal
V = 105*36.58*cos45/205 = 13.25 ft/s
3. Again by energy conservation if speed at C be V
m*V^2/2 + mgh = m*13.25^2/2
V^2 = 13.25^2 - 2gh = 13.25^2 - 2*32.17* 3
V^2 < 0
Not possible.
So it won't reach C
4. Therefore it won't reach D.