1. A device known as Atwood\'s machine consists of two masses hanging from the e
ID: 2116018 • Letter: 1
Question
1. A device known as Atwood's machine consists of two masses hanging from the ends of a vertical rope that passes over a pulley. Assume the rope and pulley are massless and there is no friction in the pulley. When the masses are of 20.1 kg and 14.9 kg, calculate the magnitude of their acceleration, a, and the tension in the rope, T. Take g = 9.81 m/s2.
2. Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 5.52 kg crate to the bottom of a steep ravine of height 26.5 meters. The 60.9 kg crewman is being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 14.7 meters above the ground, the crewman slips and the crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff. If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff.
At what speed will the crewman hit the bottom of the ravine? (Assume no air friction.)
3.A 1360-kg racecar is driving around a circular track of radius 145 m. In the instant shown in the figure, the vehicle has a forward speed of 36.5 m/s and is slowing at a rate of 5.15 m/s2. What is the magnitude of the net force acting on the vehicle at this time and the force's direction relative to the velocity?
Explanation / Answer
1.
20.1*9.8 - T = 20.1 *a
T-14.9*9.8 = 14.9*a
add two equations
50.96 = 35 a
a = 1.456 m/s^2
3.A 1360-kg racecar is driving around a circular track of radius 145 m. In the instant shown in the figure, the vehicle has a forward speed of 36.5 m/s and is slowing at a rate of 5.15 m/s2. What is the magnitude of the net force acting on the vehicle at this time and the force's direction relative to the velocity?
centripetal acceleration = V^2/R = 9.1879 m/s^2
Net acc = sqrt (9.1879^2 + 5.15^2) = 10.5328 m/s^2
Fnet = m*a = 14325.65 N
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