Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

Newton's theorem of revolving orbits wikipedia , lookup

Fundamental interaction wikipedia , lookup

Fictitious force wikipedia , lookup

Nuclear force wikipedia , lookup

Mass versus weight wikipedia , lookup

Centrifugal force wikipedia , lookup

Newton's laws of motion wikipedia , lookup

Transcript

Name Date Pd Net Force Particle Model Worksheet 1: Force Diagrams and Net Force 1. An elevator is moving up at a constant velocity of 2.5 m/s, as illustrated in the diagram below: The passenger has a mass of 85 kg. a. Construct a force diagram for the passenger. y Fnet = 0 FN = x = Fg b. Calculate the force the floor exerts on the passenger. FN = –Fg = –mg = -(85kg(-10 N/kg)) = 850 N 2. The elevator now accelerates upward at 2.0 m/s2. y a. Construct a force diagram for the passenger. Fnet FN x b. Write an equation for the vertical forces on the passenger. Fg Fnet = FN + Fg = ma c. Calculate the force the floor exerts on the passenger. Fg is -850N because it is directed downwards. Fnet FN Fg m a FN m a Fg 85kg 2.0 sm2 850N 1020N ©Modeling Instruction 2013 1 U5 Net Force – ws1 v3.1 3. Upon reaching the top of the building, the elevator accelerates downward at 3.0 m/s2. a. Construct a force diagram for the passenger. y Fnet = 0 FN x Fg b. Write an equation for the vertical forces on the passenger. Fnet = FN+ Fg = ma c. Calculate the force the floor exerts on the passenger. Fnet FN Fg m a FN m a Fg 85kg 3.0 sm2 850N 595N 4. While descending in the elevator, the cable suddenly breaks. How big is the force on the passenger by the floor? Explain your answer. The force on the passenger by the floor goes to 0 when the cable breaks because the elevator is now in "free fall". Since the elevator and the passenger are falling together, the floor no longer pushes up on the passenger. Algebraically, this is given by Fnet FN Fg m a FN m a Fg 85kg 10.0 m 2 850N 0N s 5. a. A 70 kg skydiver jumps out of an airplane. Immediately after jumping, how large is the skydiver's acceleration? Immediately after jumping, the only force on the skydiver is Fg which produces an acceleration of 10 m/s/s b. Upon reaching a downward velocity of 100 miles per hour, 300 newtons of drag resist the diver's motion. Draw a force diagram for the skydiver. How large is the skydiver's acceleration? y Fnet = Fdrag + Fg Fnet Fdrag Fnet = 300N - (70kg(10m/s/s)) Fnet = 300N - 700N = -400 N a x Fnet 400N 5.71 s m 70kg s m Fg ©Modeling Instruction 2013 2 U5 Net Force – ws1 v3.1 6. a. Draw a force diagram for a 900 kg car that exerts 5000 N of traction force on a level road while being opposed by 1000 newtons of friction and drag forces combined. y Fnet FN Fg = Ftraction x Ff = Fg b. Write a net force equation for the car. Considering only forces in the x-direction Fnet = Ftraction - F drag c. Calculate the acceleration of the car. a m Fnet (5000N 1000N) 4.44 s m 900kg s 7. The three modified Atwood's machines shown below have blocks of mass M on a frictionless surface and hanging from a string. When the blocks are released, they accelerate as they did in our lab. a. Which system has the greatest net force? Explain how you know. System C has the greatest net force. FN and Fg cancel out for the block moving horizontally. All of the "FT"s cancel out inside the systems because each string is pulling equally in both directions. The net force on each system is the force of gravity on the hanging masses. System C has the heaviest hanging masses, so the net force acting on the system is 2Mg. b. Which system has the least inertia? Explain how you know. If we define inertia as resistance to change in motion, and we know that in these systems the resistance is determined by the mass of the system, then the system with the least mass will have the least inertia. System A has the least mass. c. Determine the acceleration for each system. ©Modeling Instruction 2013 3 U5 Net Force – ws1 v3.1 In each of these systems, the Fnet is provided by the weight, mg, of the hanging block(s). System A, a m Fnet Mg g 5.0 s m 2M 2 s System B, a m Fnet Mg g 3.3 s m 3M 3 s m F 2Mg 2g System C, a net 6.7 s m 3M 3 s 8. A child takes a trip down a slide. a quantitative force diagram for the 30kg child. The frictional force is 160 N. a. Draw Fnet Ffk=160N FN=246N Fg Fg║ Fg=300N b. Write an equation for the forces on the child parallel to the slide and find the net force on the child. Fnet = Fg parallel + Ffk N Fnet mgsin55 160N 30kg 10 kg 0.819 160N 86N down the slide c. Calculate child's acceleration. a m Fnet 86N 2.87 2.9 s m 30kg s ©Modeling Instruction 2013 4 U5 Net Force – ws1 v3.1