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Aim: More Law of Inertia

Aim: More Law of Inertia. Answer Key HW 4. Do Now: Using dimensional analysis, determine what the following expression is solving for:. mass.

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Aim: More Law of Inertia

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  1. Aim: More Law of Inertia Answer Key HW 4 Do Now: Using dimensional analysis, determine what the following expression is solving for: mass

  2. A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When each has fallen 1m, they both have the same • Inertia • (B) Speed • (C) Momentum • (D) kinetic energy • (E) change in potential energy No Calculator **1 min**

  3. 2. The vertical position of an elevator as a function of time is shown above Calculator **11 min**

  4. a. On the grid below, graph the velocity of the elevator as a function of time The velocities are equal to the slopes from the position-time graph

  5. b. i. Calculate the average acceleration for the time period t = 8 s to t = 10 s.

  6. ii. On the box below that represents the elevator, draw a vector to represent the direction of this average acceleration aavg

  7. c. Suppose that there is a passenger of mass 70 kg in the elevator. Calculate the apparent weight of the passenger at time t = 4 s. • The apparent weight is the normal force • At t = 4 s, there is constant velocity, therefore: • ΣF = 0 • N – mg = 0 • N = mg • N = (70 kg)(9.8 m/s2) • N = 686 N

  8. Calculator allowed **7 minutes** 3. Two 10‑kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown above. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the Ieft‑hand box and the plane are 0.15 and 0.30, respectively. You may use g = 10 m/s2, sin 60° = 0.87, and cos 60° = 0.50.

  9. a. What is the tension T in the string? ΣF = 0 T = mg T = (10 kg)(10 m/s2) T FN b. On the diagram to the right, draw and label all the forces acting on the box that is on the plane FF Fg FF + F║= T FF = T - F║ FF = T – mg(sin60°) FF = 100 N – (10 kg)(10 m/s2)(0.87) FF = 13 N c. Determine the magnitude of the frictional force acting on the box on the plane.

  10. Calculator allowed **15 minutes** 4. An empty sled of mass 25 kg slides down a muddy hill with a constant speed of 2.4 m/s. The slope of the hill is inclined at an angle of 15° with the horizontal as shown in the figure above. a. Calculate the time it takes the sled to go 21 m down the slope.

  11. b. On the dot below that represents the sled, draw and label a free-body diagram for the sled as it slides down the slope. c. Calculate the frictional force on the sled as it slides down the slope.

  12. d. Calculate the coefficient of friction between the sled and the muddy surface of the slope.

  13. e. The sled reaches the bottom of the slope and continues on the horizontal ground. Assume the same coefficient of friction. i. In terms of velocity and acceleration, describe the motion of the sled as it travels on the horizontal ground. The velocity of the sled decreases with a constant acceleration

  14. ii. On the axes below, sketch a graph of speed v versus time t for the sled. Include both the sled’s travel down the slope and across the horizontal ground. Clearly indicate with the symbol tl the time at which the sled leaves the slope.

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