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Timer #1 (group member 2)Timer #2 (group member 3) Trial #123123Average Larger Car 1/2 track 0.720.910.710.850.800.890.81 Larger Car full track 1.001.06.

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Presentation on theme: "Timer #1 (group member 2)Timer #2 (group member 3) Trial #123123Average Larger Car 1/2 track 0.720.910.710.850.800.890.81 Larger Car full track 1.001.06."— Presentation transcript:

1 Timer #1 (group member 2)Timer #2 (group member 3) Trial #123123Average Larger Car 1/2 track 0.720.910.710.850.800.890.81 Larger Car full track 1.001.06 1.011.031.05 1.04 Smaller Car 1/2 track 0.780.720.690.680.710.75 0.72 Smaller Car full track 1.001.071.031.050.990.98 1.02 SAMPLE DATA: Yours should be different The yellow column is you AVERAGE TIME

2 Timer #1 (group member 2) Timer #2 (group member 3) Trial #123123Average Larger Car 1/2 track 0.720.910.710.850.800.890.81 Larger Car full track 1.001.06 1.011.031.051.04 Smaller Car 1/2 track 0.780.720.690.680.710.750.72 Smaller Car full track 1.001.071.031.050.990.981.02 Length section #2 = 53.4 cm Length of #2 and #3 = 106.8 cm In order to calculate the AVERAGE SPEED that you are asked for in the first 4 questions, you need to take the distance of that portion of the track (seen under your data table) and divide it by the appropriate AVERAGE TIME found in your data table.

3 Once you know the average speed for each of the 4 required problems THEN you can calculate the Accelerations using the correct formula. Final Speed – Initial Speed Time Paper Box Track Section #1 Track Section #2 Track Section #3 You are being asked to find the acceleration of the car on the Second half of the track. Speed here is FINAL Speed here is INITIAL You will take the calculated speeds for the areas shown in red above and plug them into the formula. Then you will need to figure out the time to plug in By subtracting the AVERAGE TIME of ½ track from the AVERAGE TIME of the full track in your data table Answer the analysis questions using your calculated speeds.

4 A box with a weight of 40 N is sitting on a table, are there forces acting on the box?

5 40 N The answer is YES! The force of the box pushing on the table is BALANCED by the force of the table holding up the box. If the table were not exerting an equal force back on the box the box would fall through the table. The forces exerted by objects can be represented by arrows. The size of the arrow indicates the size of the force as well as the direction

6 40 N Let’s add another force to the system: a person decides to push the box with a force of 50 N to the left (the black arrow represents this new force) Which direction would the box move? 50 N

7 40 N To the LEFT! 50 N The object will always move in the direction of the stronger, unbalanced force. What if there were another person on the other side of the box pushing to the Right with a force of 50 N as well? Motion of the box With 50 N of force

8 40 N 50 N The box would remain still Now the forces are equal and opposite again so the box will not move. The system has been balanced with NET FORCE of zero.

9 40 N 50 N The person on the right begins to get tired and his force decreases to 40 N, what will happen to the box?

10 40 N 50 N The box would move to the RIGHT! The person on the left now has more force so the box moves to the right. When forces are unbalanced the object will move in the direction of the stronger force (the direction that it is pushing). The forces are subtracted in order to find the NET FORCE acting on the box. Motion of the box with a net force of 10 N

11 40 N What if the other person decided to help push the box to the left with a force of 20 N? 50 N

12 40 N The box would move to the left with 70 N of force! 50 N 20 N When forces are acting in the same direction they are added together. The combination of forces is called the NET FORCE Notice that the 40 N force of the box does not get included because it was balanced out by the force of the table. Motion of the box With a net force of 70 N

13 Let’s Practice This I know you’re excited!

14 According to the force arrows shown in this picture (red represents the boats force to move forward, blue represents the force of the river) will the person in the boat get closer to the waterfall?

15 The person’s force in this picture is much smaller than the force of the river That is moving in the opposite direction. The canoe will drift back wards Away from the waterfall. NO, the canoe will move further away

16 Green = Force exerted by river Yellow = Force exerted by the man rowing the boat In this example the river is pushing on the boat with a force of 20 N and the man rowing the boat exerts a force of 10 additional Newtons. What is the Net Force exerted on the boat?

17 Green = Force exerted by river Yellow = Force exerted by the man rowing the boat If the man stopped rowing would the boat continue to move? What would be the Net Force exerted on the canoe? Net force = 30 N

18 Green = Force exerted by river Yellow = Force exerted by the man rowing the boat The boat would continue to move because the river’s force is still unbalanced. What would the rower have to do in order to get the boat to stop moving? Yes, it would move with a net force of 20 N.

19 Green = Force exerted by river Yellow = Force exerted by the man rowing the boat He would need to row the other direction with a force of 20 N in order to balance the forces and stop the boat. What would happen if the man told his wife to stop lounging around and help paddle?

20 The force of gravity acting on the mass of the canoe may eventually overcome the force that the man is able to exert while carrying it. If this happens, the force of gravity will win and the man will drop the canoe Red = force of canoe due to gravity Dark blue = forced exerted by the man *the force of the woman screaming at the man after being told to paddle has not been pictured

21 Complete the practice problems located on your paper


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