Aim: More conservation of Energy Do Now: A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground? ΔKE = ΔPE ΔKE = mgΔh ΔKE.

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Presentation transcript:

Aim: More conservation of Energy Do Now: A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s2)(10 m) ΔKE = 196 J HW #10 Answer Key

1. A ball thrown vertically downward strikes a horizontal surface with a speed of 15 meters per second. It then bounces, and reaches a maximum height of 5 meters. Neglect air resistance on the ball. a.What is the speed of the ball immediately after it rebounds from the surface? b. What fraction of the ball's initial kinetic energy is apparently lost during the bounce? Calculator **5 minutes**

2. A 0.10-kilogram solid rubber ball is attached to the end of an 0.80-meter length of light thread. The ball is swung in a vertical circle, as shown in the diagram above. Point P, the lowest point of the circle, is 0.20 meter above the floor. The speed of the ball at the top of the circle is 6.0 meters per second, and the total energy of the ball is kept constant. Calculator **15 minutes**

a.Determine the total energy of the ball, using the floor as the zero point for gravitational potential energy. E T = U + K E T = mgh + ½ mv 2 E T = (0.1 kg)(9.8 m/s 2 )(1.8 m) + ½ (0.1 kg)(6 m/s) 2 E T = 3.6 J

b. Determine the speed of the ball at point P, the lowest point of the circle.

c. Determine the tension in the thread at i. the top of the circle

ii. the bottom of the circle.

The ball only reaches the top of the circle once before the thread breaks when the ball is at the lowest point of the circle. d.Determine the horizontal distance that the ball travels before hitting the floor. y = v o t + ½ at 2 y = ½ gt 2 t 2 = 2y/g t 2 = 2(0.2 m)/9.8 m/s 2 t 2 = 0.04 s 2 t = 0.2 s v x = x/t x = v x t x = (8.2 m/s)(0.2 s) x = 1.6 m

P Calculator **17 minutes** 3.

ii. Calculate the value v max of this maximum speed

b. Calculate the speed v B of the car at point B.

FgFg N F g = mg F g = (700 kg)(9.8 m/s 2 ) F g = 6860 N

Flatten out the track on either side of the loop so the bottom of the loop is on the ground, thus lowering point B. The work done by friction will reduce the total mechanical energy available at point B, so if the kinetic energy at point B is to remain the same, the potential energy at that point must be reduced.