1. <ul><li>Read Investigation 2.3 beginning on page 157. </li></ul><ul><li>Start a new page in your lab notebook. Title it “Newton’s Second Law of Motion” – don’t forget your table of contents! </li></ul>
2. <ul><li>Predict what will happen when a cart is pushed with a constant force. Write an “if, then” statement in your lab book. </li></ul><ul><li>Predict what will happen when you use the same amount of force but have a cart with less mass. Write an “if, then” statement in your lab book. </li></ul>
3. <ul><li>Predict what will happen when you push a cart with increased mass with the same force. Write an “if, then” statement in your lab book. </li></ul>
4. <ul><li>Make a list of materials you will use in today’s lab. Put this list in your lab book. </li></ul>
5. <ul><li>Today you will need: </li></ul><ul><ul><li>calculator </li></ul></ul><ul><ul><li>spiral </li></ul></ul><ul><ul><li>pen or pencil </li></ul></ul><ul><li>Today we will: </li></ul><ul><ul><li>review what we learned Friday </li></ul></ul><ul><ul><li>take some notes </li></ul></ul><ul><ul><li>complete a worksheet </li></ul></ul>
6.
7. <ul><li>There are two types of observations we can make during an investigation </li></ul><ul><ul><li>Qualitative – qualities of objects, events , or processes : something smells spicy, tastes sweet, feels slippery </li></ul></ul><ul><ul><li>Quantitative – observations based on counting or measuring: the temperature, the distance, the speed </li></ul></ul>
8. <ul><li>What type of measurements did we make in this investigation? </li></ul>
9. <ul><li>Newton’s Second Law can be expressed with an equation: </li></ul><ul><li>Which can be re-arranged to isolate Force like this: </li></ul>
10. <ul><li>Force is measured in Newtons </li></ul><ul><li>A Newton = the amount of force needed to accelerate a 1 kilogram mass at a rate of 1 m/s 2 </li></ul><ul><li>Since F = m . a </li></ul><ul><li>1 N = 1 kg . m/s 2 </li></ul>
11. <ul><li>If there is acceleration (speed up, slow down, change direction), there is an unbalanced force. </li></ul><ul><li>remember: another name for unbalanced force is net force </li></ul>
12. <ul><li>Look at the relationship between force, mass, and acceleration in this equation: </li></ul><ul><li>To keep the same acceleration, what must force do if mass is increased? </li></ul><ul><ul><li>force must increase just as much </li></ul></ul>
13. <ul><li>To be more specific: </li></ul><ul><ul><li>If acceleration stays constant and the mass of the object doubles, what must the force be doing? </li></ul></ul><ul><ul><li>doubling </li></ul></ul>
14. <ul><li>Turn in your book to page 163 </li></ul>
15. <ul><li>IN YOUR SPIRAL: </li></ul><ul><li>Describe the motion of the object this graph depicts </li></ul><ul><li>Using F = ma, calculate the mass of the object </li></ul><ul><li>Collaborate with your table partners – make sure answers agree </li></ul>
16. <ul><li>“Sig Figs” </li></ul><ul><li>When you perform calculations that use measurements ( quantitative ), you need to express the results of your calculations in a way that makes sense of the precision of the measurements you used. </li></ul>
17. <ul><li>In other words, if you measured your distance as 2.3 centimeters and the time as 1.1 s, you wouldn’t report the speed as 2.0909 cm/s because it implies a precision you didn’t have with your measurements </li></ul>
18. <ul><li>So, how do you figure out how many figures to use? </li></ul><ul><li>There are five rules </li></ul><ul><ul><li>1. All non zero numbers are significant figures </li></ul></ul><ul><ul><li>So…if the measurement is 125.5 m, how many sig figs are there? </li></ul></ul><ul><ul><li>four </li></ul></ul>
19. <ul><li>2. A zero at the end of a decimal number IS significant. So, if the measurement is given as 1.50N, the zero IS significant </li></ul><ul><ul><li>How many sig figs in 31.40 g? </li></ul></ul><ul><ul><li>four </li></ul></ul>
20. <ul><li>3. A zero between nonzero digits is significant. In the measurement 405 km, the zero is significant. </li></ul><ul><ul><li>How may sig figs in 307.89 g? </li></ul></ul><ul><ul><li>five </li></ul></ul>
21. <ul><li>4. A zero at the beginning of a decimal point is not significant. In the measurement 0.027kg, the zeros are not significant. That measurement has two sig figs. </li></ul><ul><ul><li>How many sig figs in the measurement 0.0306 g? </li></ul></ul><ul><ul><li>three </li></ul></ul>
22. <ul><li>5. In a large number without a decimal point, the zeros are not significant. In the measurement 2000 km, there is only one significant figure. </li></ul><ul><ul><li>how many sig figs in the measurement 500m? </li></ul></ul><ul><ul><li>one </li></ul></ul><ul><ul><li>how many sig figs in 501 g? </li></ul></ul><ul><ul><li>three </li></ul></ul>
23. <ul><li>Adding & Subtracting </li></ul><ul><ul><li>the final result should have the same number of decimal places as the measurement with the fewest decimal places </li></ul></ul>
24. <ul><li>Multiplying and Dividing </li></ul><ul><ul><li>the result should have no more significant digits than the factor having the fewest number of significant digits </li></ul></ul>
25. <ul><li>27.09 km has how many significant figures? </li></ul><ul><ul><li>four </li></ul></ul><ul><li>3600 g has how many significant figures? </li></ul><ul><ul><li>two </li></ul></ul><ul><li>54.009 has how many significant figures? </li></ul><ul><ul><li>five </li></ul></ul>
26. <ul><li>10.2 + 201 = ? </li></ul><ul><ul><li>211 – you can’t say it’s 211.2 because that would be four significant figures and both numbers you are adding have three significant figures </li></ul></ul>
27. <ul><li>0.02991 x 10.0 = ? </li></ul><ul><ul><li>0.299 – there are three significant figures in 10.0, so that’s how many significant figures you should have in your answer. </li></ul></ul>
28. <ul><li>We already know that the formula for the second law of motion is F = ma </li></ul><ul><li>We can use this formula to calculate the force of weight </li></ul><ul><li>Weight = mass x acceleration gravity </li></ul>
29. <ul><li>What happens when an unbalanced force acts on an object? </li></ul><ul><ul><li>the object accelerates </li></ul></ul><ul><li>When two forces act on an object at the same time, the direction as well as the magnitude of the force determine the motion of the object </li></ul>
30. <ul><li>If the two forces are in the same direction, the sum of the forces (net force) will cause a larger acceleration than either force on its own </li></ul>
31. <ul><li>If the two forces are in opposite directions then the net force could be zero – in which case there would be no acceleration </li></ul><ul><li>OR </li></ul><ul><li>the sum of the two forces will result in a net force in one direction </li></ul>
32. <ul><li>Draw a free body diagram that you think would represent two forces in opposite directions that would result in no net force (and thus, no acceleration) </li></ul>
33. <ul><li>Draw a free body diagram with two forces in opposite directions that would result in a net force (and acceleration) </li></ul>
34. <ul><li>It’s really, really important to remember: </li></ul><ul><li>a net force results in acceleration </li></ul><ul><li>BUT </li></ul><ul><li>no net force does not mean no motion – an object will move at a constant speed in a straight line with no net force!!! </li></ul>
35. <ul><li>What if the forces aren’t in line with each other? </li></ul>
36. You add the vectors using tip-to-tail and then use the Pythagorean theorem to solve a 2 + b 2 = c 2 (40 N) 2 + (60 N) 2 = c 2 1600 N 2 + 3600 N 2 = c 2 5200 N 2 = c 2 c = 72 N
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