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Hooke’s law This lesson introduces forces from springs and Hooke's law: F = −kx. The presentation begins by describing types of springs and distinguishing between a spring’s free length and its deformation x. The meaning of the spring constant is explained. In a brief hands-on activity, students use a spring scale to measure the spring constant of rubber bands. Students then perform an investigation in which they collect and graph data of the force and deflection for both a stiff spring and a weaker spring, and draw inferences from their graphs.

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Objectives Calculate the force from a spring when given its spring constant and deflection. Calculate a spring constant given the required force and deflection. The lesson objectives describe what a student should know and be able to do.

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Assessment A certain spring with a free length of 10.0 cm has a spring constant of k = 500 N/m. How much force does the spring exert if it is extended to a length of 12.0 cm? What is the required spring constant for a spring to support a 10.0 kg mass while deflecting only 14 cm? This first assessment is keyed to the first objective: calculate the force from a spring when given its spring constant and deflection. The second assessment is keyed to the second objective: calculate a spring constant given the required force and deflection. These assessments will be repeated at the end of the lesson, followed by the answers

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**Physics terms Hooke’s law spring constant extension compression**

deflection deformation New or key vocabulary words are presented on this slide.

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**Equations Hooke’s law:**

The force exerted by a spring is the negative of the product of the spring constant multiplied by the deformation of the spring. New equations are presented on this slide. Translating mathematical equations into spoken English is a useful exercise for helping students understand that math is a kind of language.

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Springs The free length of a spring is its length without any external forces applied. Fs

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Springs The free length of a spring is its length without any external forces applied. Fs The deformation x is the length in meters that the spring is extended (+x) or compressed (-x). Students need to be careful in distinguishing the deformation of a spring, x, from its length.

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Springs The deformation x of the spring depends on how much force is exerted to stretch it or compress it. Fs Students need to be careful in distinguishing the deformation of a spring, x, from its length.

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Investigation In Investigation 5B you will measure the force and deformation of a stretched spring. How are these variables related? The investigation is found on page 150. This investigation appears on page 150 of the textbook.

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**Investigation Part 1: Extension and restoring force**

Set up the equipment using the looser spring. With the spring scale attached, mark the equilibrium position of the bottom of the spring on the ruled paper as “0 N.” Pull down the spring scale by 1 N to extend the spring. Mark the new location of the end of the spring on the ruled paper with a label “1 N.”

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**Investigation Part 1: Extension and restoring force**

Pull down the spring for forces of 2, 3, 4, and 5 N. Mark the position of the end of the spring and label each mark with the force. Remove the paper and measure the distance x (in meters) of each point from equilibrium. Graph your data with deformation x on the horizontal axis.

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**Investigation Questions for Part 1 What is the slope of your graph?**

What physical quantity is represented by the slope of your graph? Why? In steps 4 and 5, what were the independent, dependent, and controlled variables? If students have any difficulty here, it may help to stop and go through the slides at the end of the investigation in which the spring constant is calculated from a graph.

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**Make inferences from the data**

Questions for Part 1 Use your graph to determine the force the spring would exert at other deformations.

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**Investigation Part 2: Stiff and loose springs**

Now substitute a “stiff” spring for the “loose” one. Then set up the experiment as before. Repeat the steps of stretching the spring scale to different forces. Record and graph your data.

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**Investigation Questions for Part 2**

When you stretch the stiff spring by hand, how does it feel or respond that is different from the loose spring? How does the extension of the stiff spring compare to that of the loose one for the same applied force?

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**Hooke’s law Deformation (m) Force (N) Spring constant (N/m)**

Hooke’s law describes an “ideal” spring. It is a good approximation for real springs. Deformation (m) Force (N) Spring constant (N/m) Ask the students: “What happens to the force if I double the stretch? Triple the stretch?” Hooke’s law is not a “law” in the same way that Newton’s laws are fundamental rules of the physical world. Hooke’s law is an empirical description of the behavior of most springs. It is generally a very good approximation for many springs within their elastic range. Tightly wound extension springs obey the law very well after they have been stretched an initial amount. Notice the units: The deformation must be in meters!

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**What is a spring constant?**

The spring constant k is a property of the spring itself. The spring constant tells you how much force F is needed to deform the spring a distance x. The units of k :

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**What is a spring constant?**

The spring constant tells you how stiff the spring is. Stiff springs have high spring constants. Weak springs have low spring constants. Tell the students: “The spring constant in essence tells you how many newtons it takes to stretch it one meter.” (Of course it may not be able to really stretch that far!) Example: For a stiff spring with k = 1000 N/m, it takes 1000 N to stretch it one meter. For a weak spring with k = 10 N/m, it takes only 10 N to stretch it one meter. Which of these springs has the higher spring constant?

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Test your knowledge Which has a higher spring constant: the rubber band or the spring in a car suspension? This example involves springs in the students’ everyday lives.

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**The car’s suspension spring has a higher spring constant.**

Test your knowledge Which has a higher spring constant: the rubber band or the spring in a car suspension? The car’s suspension spring has a higher spring constant. How do you know?

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**The car’s suspension spring has a higher spring constant.**

Test your knowledge Which has a higher spring constant: the rubber band or the spring in a car suspension? The car’s suspension spring has a higher spring constant. How do you know? It is much stiffer—it takes more newtons of force to stretch or compress it. The suspension spring must be stiff. Otherwise it will deflect to much when subjected to the large forces from a car or truck bouncing over a bumpy road.

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**Measure k of a rubber band.**

Materials: Ruler, spring scale, #33 rubber band Hook one end of the rubber band over the zero end of a ruler. Straighten the rubber band to its free length. Typical values for the spring constant of the #33 rubber band will be about 60 N/m. As students calculate an answer, write some of these on the for reference. Note on signs: The external force the student exerts on the spring is equal and opposite to the force the spring exerts on the student. These are an action/ reaction pair. Therefore the negative sign cancels out.

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**Measure k of a rubber band.**

Materials: Ruler, spring scale, #33 rubber band Stretch the rubber band 0.10 m (10 cm). Calculate the spring constant: Typical values for the spring constant of the #33 rubber band will be about 60 N/m. As students calculate an answer, write some of these on the for reference. Note on signs: The external force the student exerts on the spring is equal and opposite to the force the spring exerts on the student. These are an action/ reaction pair. Therefore the negative sign cancels out. x = 10 cm

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**Measure a different rubber band**

Materials: Ruler, spring scale, #16 rubber band A #16 rubber band is thinner than a #33. Repeat the measurements and calculate a spring constant for the #16 rubber band. Predict: Will it have a higher or lower spring constant than a #33 band? The #16 rubber band will be less stiff. Typical values for the spring constant of the #16 rubber band will be around 20 N/m to 40 N/m.

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Springs in parallel Materials: Ruler, spring scale, two #16 rubber bands Combine 2 rubber bands in parallel (side by side ). Repeat the measurements and calculate the spring constant for this parallel combination. Predict: Will the combination have a higher or lower spring constant than one alone? The parallel combination will be stiffer by a factor of about two, so typical values for combination will be around 40 N/m to 80 N/m. (If the rubber bands were really identical ideal springs, the parallel combination would have exactly twice the spring constant of the single rubber band.)

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**Springs in series Materials: Ruler, spring scale, two #16 rubber bands**

Combine 2 rubber bands in series by looping one onto the other one. Repeat the measurements and calculate the spring constant for this series combination. Predict: Will the combination have a higher or lower spring constant than one alone? The series combination will be less stiff, with about half the spring constant of the single rubber band. (If the rubber bands were really identical ideal springs, the series combination would have exactly half the spring constant of the single rubber band.)

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**Hooke’s law Force (N) Spring constant (N/m) Deformation (m)**

What is the meaning of the minus sign in Hooke’s law? Force (N) Spring constant (N/m) Deformation (m)

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**Hooke’s law Force (N) Spring constant (N/m) Deformation (m)**

What is the meaning of the minus sign in Hooke’s law? The force exerted by the spring is in the opposite direction from the deformation! Force (N) Spring constant (N/m) Deformation (m) The negative sign indicates that the spring force is a “restoring force”. When you deflect a spring, a force arises that tries to restore the system to its equilibrium position.

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**The meaning of the minus sign**

Fs When the deformation x is in the positive direction, the force Fs exerted by the spring is in the negative direction. Point out that the vector arrows for F and x point in opposite directions in the diagram.

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**The meaning of the minus sign**

Fs When the deformation x is in the negative direction, the force Fs exerted by the spring is in the positive direction.

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Exploring the ideas Click this interactive calculator on page 149.

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**Engaging with the concepts**

Set the force to zero and solve for the deformation. What happens? 10 Deformation

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**Engaging with the concepts**

Set the force to zero and solve for the deformation. What happens? At zero force the spring is at its free length and the deformation is zero. 10 Deformation When the force is zero, it is at its free length. The rightmost end of the spring lines up with zero, and the deformation is zero.

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**Engaging with the concepts**

Enter a deformation of m. What is the force? 10 0.50 Force

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**Engaging with the concepts**

Enter a deformation of m. What is the force? -5 N Why is it negative? -5 10 0.50 Force

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**Engaging with the concepts**

Enter a deformation of m. What is the force? -F +x -5 N Why is it negative? -5 10 0.50 Because the spring force F pulls in the opposite direction of the stretch, x. Force

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**Engaging with the concepts**

Enter a deformation of m. What is the force? 10 0.50 Force

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**Engaging with the concepts**

Enter a deformation of m. What is the force? -x +F +5 N The force created by the spring now pushes to the left. 5 10 0.50 Force

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**Calculating the spring constant**

Force vs. deflection. Examine this graph of force versus deflection. What is the slope of this graph?

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**Calculating the spring constant**

Force vs. deflection. Examine this graph of force versus deflection. What is the slope of this graph? rise What physical quantity is represented by the slope? run Point out the that calculation is based on the best fit line, not on individual data points. The slope of the line is the same at all points. The slope was calculated using a section of the line that intersected the gridlines for improved accuracy.

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**Calculating the spring constant**

Force vs. deflection. Examine this graph of force versus deflection. What is the slope of this graph? rise What physical quantity is represented by the slope? run The slope is the spring constant, k: the force the spring exerts per meter of stretch.

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Extending Hooke’s law Consider this question: A bowling ball rests on a table. The table pushes up on the ball with a normal force exactly equal to the ball’s weight. How does the table “know” how much force to push with? Hooke’s law is not just a description of springs. It is a useful law for understanding any system that has some degree of elasticity, from a table top to the beams of a bridge or skyscraper, to the vibrating elements of a musical instrument.

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Real objects Real objects deflect under applied forces, just like springs.

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Real objects The table acts like a spring. Its “spring constant” determines how much the table deflects under any given force. It continues to deflect until forces come to equilibrium.

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Real objects For small deflections, the relationship is approximated by Hooke’s law.

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Types of springs Real springs come in many different types. Hooke’s law can be used to describe the force exerted by all kinds of springs. Hooke’s law is an approximation that is useful for all kinds of springs. Show the students a variety of real springs if you have a collection available. Springs available for classroom use are usually extension springs.

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Assessment A certain spring with a free length of 10.0 cm has a spring constant of k = 500 N/m. How much force does the spring exert if it is extended to a length of 12.0 cm? This first assessment is keyed to the first objective: calculate the force from a spring when given its spring constant and deflection. The answer appears on the next slide.

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**Assessment F = -kx = - (500 N/m) (0.02 m) = -10 newtons**

A certain spring with a free length of 10.0 cm has a spring constant of k = 500 N/m. How much force does the spring exert if it is extended to a length of 12.0 cm? What is the required spring constant for a spring to support a 10 kg mass while deflecting only 14 cm? F = -kx = - (500 N/m) (0.02 m) = -10 newtons The second assessment is keyed to the second objective: calculate a spring constant given the required force and deflection. The answer appears on the next slide.

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**Assessment F = -kx = - (500 N/m) (0.02 m) = -10 newtons**

A certain spring with a free length of 10.0 cm has a spring constant of k = 500 N/m. How much force does the spring exert if it is extended to a length of 12.0 cm? What is the required spring constant for a spring to support a 10 kg mass while deflecting only 14 cm? F = -kx = - (500 N/m) (0.02 m) = -10 newtons Weight mg = (10 kg)(9.8 N/kg) = 98 N down Since the mass is at rest with Fnet = 0 N, then Fspring must be 98 N up. From Hooke’s law, k = - F/x = -(98 N) / (-0.14 m) = 700 N/m

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